w '*&*" GP^^SO W><^ ^C^fcJ^^^Q^^E^! 7 3 ^ fe^i ( :v^l?^fe p| $!!: .j^^&i' riH p^ ^IrW^ ELECTRICAL ENGINEERING FIRST COURSE McGraw-Hill BookCompany Electrical World The Engineering and Mining Journal Engineering Record Engineering News Railway Age Gazette American Machinist Signal Hngin where G is the conductance of a circuit of resistance R. K Likewise the reciprocal of resistivity, called conductivity, is often used. The resistance of a wire at any temperature t, when its resistance at any other temperature is known can be calculated by the following equation R t = Rt.ll + a tl (t - t,)} When ti = 0C. then R t = R (l + aj) where R t is the required resistance at any temperature, t, R in this case is the resistance at 0, and a is a constant, called the temperature coefficient. For copper, a = 0.004 (approximately) when t is given in Centigrade degrees. At any other temperature the value of a is: 1 234.5 + t where t is the temperature in degrees C. Since a depends upon the temperature, in all calculations in- volving a its value is calculated for that temperature at which the resistance is known. Knowing the resistance Ri at a temperature ti the resistance Rz at temperature tz is thus accurately determined from the following relation: 234.5 + t, 234.5 + UNITS 3 TABLE I Table I gives approximately the temperature coefficients and resistivities in ohms per centimeter cube of some of the more common electrical con- ductors at ordinary temperature. Conductor Temp, coefficient a Resistivity Aluminium 0.0042 2.9 X 10~ 6 Carbon 00052 720 X 10~ Copper 0.004 1.6 X 10~ 6 German silver 0.00027 20.9 X 10~ 6 Iron 0.0046 9 7 X 10~ 6 Nickel 0.0062 12.4 X 10~ Platinum . . . 0.0036 9 X 10- Silver . 0.004 1.5 X 10~ 6 Tungsten 0.005 5 X 10- Development of Ohm's Law. According to OHM'S law the current in a circuit at any instant is equal to the potential differ- ence divided by the resistance, or, 7 = R' Obviously, where a number of resistances are in series, the total resistance is the sum of the individual resistances, or, Rtotal ~ 2r = TI + TZ + 7*3 -f- . . . Two Resistances in Parallel. To find the total current 7, and the currents 7i, Iz in the resistances r\ and r 2 , when a potential difference E is applied (Fig. 1). / By OHM'S law, Tjl Tjl r Hi j and FIG. 1. E E To find a single resistance, r , which shall be the equivalent of ri and r 2 in parallel, evidently ' 4 ELECTRICAL ENGINEERING Whence, r a = r 2 Having two resistances in parallel, in series with a third resist- ance (Fig. 2), to find the combined resistance. Let the combined resistance of r*i and r 2 be r . Then r = - T\ -J- 7*2 The condition is, then, that of two re- i : Sr i r sistances r and r 3 in series and the total j>*\ r >i 3 n resistance R = r + r 3 . + J Hence r * / - - - -L """ yv FIG. 2. 72 r + r 3 To find /i and /2. It is evident that / = 7 3 . Knowing 7 3 and r 3 , we may at once determine E$ which is the potential difference, or drop, across r 3 . Thus, by OHM'S law, E z = / 3 r 3 . It is evident that the potential difference E , across r\ and T* is E E$. ' T _ E . T - ^ - Tt> /2 ~ r 2 General Solution of a Network by Kirchoff's Laws. In cir- cuits or networks of a more complicated nature in which the resistances and electromotive forces are known, the currents in the various branches may be calculated by the application of KIRCHOFF'S laws which may be stated as follows: Law I. The algebraic sum of all the currents flowing toward a branch point is equal to zero. _ Law II. The algebraic sum of all the e.m.fs. acting around a closed circuit is equal to the sum of the products, ri, around the mesh. Or the impressed e.m.f. is equal to the sum of all e.m.fs. consumed by the resistances. For example, let the circuit be as shown in Fig. 3 where arrows represent arbitrarily chosen direc- tions of current. For the points A, B, C, D, applying Law I, equations may be written: UNITS A. B. C. D. - i - is - i* - t' 4 = ii + 12 - i s = i 4 + *B - i = 0. (1) (2) (3) (4) Applying Law II, where the short arrow represents the direc- tion of the e.m.f., to the meshes (a) e, r 3 , r 4 , (b) e, ri, r 5 , (c) n f r, r 3 , (d) r 2 , r 5 , r 4 , always keeping an arbitrarily chosen counter- clockwise direction, we have, (a) ri + r 3 i 8 + r 4 i 4 = e (5) (b) ri + nil + r b i b = e (6) (c) nil - r,i 2 - r 3 i 3 = (7) (d) r 2 iz + r 6 i 8 - r 4 i 4 = (8) There is one extra equation in each group as there are only six unknown quantities, i, ii, i 2 , is, i 4 , is. In calculating the resistance of more or less complex circuits it is helpful to remember that current does not flow between points of the same potential. If, in Fig. 3, there is no difference of potential between points B and C there will be no current in the branch r 2 . PROBLEMS Problem 1. If the resistivity (resistance of a cubic centimeter between parallel faces at 0C.) of copper is 1.6 X 10~ 6 ohm, (a) show that the resist- ance of an inch cube of copper is 0.63 X 10~ 6 ohm; (b) show that if the temperature coefficient, a. = 0.004, the resistance of a centimeter cube at 20C. is 1.73 X 10~ 6 ohm; (c) show that __ B the temperature coefficient per degree Fahrenheit is 0.0022. _ FIG. 4. FIG. 5. Problem 2. If a wire be connected across the terminals of a source of constant e.m.f., a current will flow. Will this current increase, decrease, or remain constant as time goes on, and why? Problem 3. Deduce the equation for the equivalent resistance of three resistances connected in parallel. Problem 4. Find the line current 7, and the voltage across r 3 in the circuit, shown in Fig. 4. E = 100 volts, r : = 1, r 2 = 2, r 3 = 3. Problem 5. Let the outline of a cube, Fig. 5, consist of resistances, each 6 ELECTRICAL ENGINEERING edge being 1 ohm. Prove that the total resistance between A and B is ^f 2 ohm; between A and C is % ohm; between A and D is % ohm. Effects of Current in a Wire. When a current is set up in a wire three effects may be noted, namely: (1) the wire gets warm, (2) a compass needle placed near the wire is deflected, and (3) when the voltage is high enough bits of paper may be attracted. The amount of energy delivered through the wire does not bear a relation to any one of these effects, but if the second and third effects are multiplied together, or, as commonly expressed, if the strength of the magnetic and electric fields are multiplied to- gether the product is a value which is proportional to the amount of energy transmitted through the wire per second, or to the power. Thus we may write, P = kei where P is the power and k is a constant, k is unity when e, which is proportional to the strength of the electric field is ex- pressed in volts, i t which is proportional to the strength of the magnetic field is expressed in amperes, and P is in watts. The first effect, that is, the production of heat is due to con- sumption of energy in the wire due to its resistance. The second effect is due to the setting up of a magnetic field about the wire by the current. The third effect is due to the setting up of an electric or electro-static field in the region about the wire by the difference of potential between the wire and other points in space. Power. In a given circuit, then, P = El = IE X / = PR in which E is the total e.m.f., I the current, and R the total re- sistance of the circuit. This relation, known as JOULE'S law, is very important, as it shows that the power is proportional to the square of the current strength and to the first power of the resistance. The heat developed by this power depends upon the duration of the current, and is expressed in joules. Thus, heat energy = Elt = I 2 Rt joules, where E is in volts, 7 in amperes, and t in seconds (the current and voltage being assumed constant during time t). r* In general, the energy converted to heat is W = I i 2 rdt. Jti Problem 6. Prove that if the current is represented by equation i I sin ut UNITS 7 T the energy per cycle is W = 7 2 r -^ where T is the time of a complete cycle. 4 W Pr The average power is then -7=- = -=- j. z (/I 2 /3 2 \ ~o~ + ~o~) r when i = /i sin co + / 3 sin (3J + a). Heat Units. The practical heat units most frequently dealt with are the British thermal unit (B.t.u.), and the large and small calories (C. and c.). One B.t.u. is the energy required to raise the temperature of 1 Ib. of water 1F. 1 B.t.u. = 1.055 kw. sec. One large calorie is the energy required to raise the temperature of 1 kg. of water 1C. 1 C. = 4.2 kw. sec. One small calorie is the energy required to raise the temperature of 1 gram of water 1C. 1 c. = 0.0042 kw. sec. Problem 8. A 16-cp. lamp which consuntes 3 watts per cp. is immersed in a quart of water at 20C. Assuming no loss of heat, (a) what will the temperature of the water be after 2 min. ? (6) How long would it take to evaporate the water? Solution. (a) Temp, will be 20 + C. rise. C.rise = ^ W ^ Xqt.inlkg. 2 kw. sec. = X 16 X 2 X 60 = 5.76 qt. per kg. = 1.057 f\ 7 A .'. C. rise = -~- X 1.057 = 1.45. Temp, after 2 min. = 21.45C. (6) Time to evaporate = time to raise .to boiling + time required to furnish latent heat of vaporization. Time required to boil 1 qt. = time to raise 1 qt. 1 X (100 - 20) = ~ X 80 = 110.3 min. Time required to evaporate = calories required to evaporate -* calories per min. supplied by lamp. = 742 min. .'. Total time required = 110.3 + 742 = 852.3 min. = 14 hr. 12 min. 8 ELECTRICAL ENGINEERING Problem 9. Transform problem 8 into F. and B.t.u. Problem 10. If electric energy costs lOc. per kw. hr., how much would it cos,t to prepare a hot bath by electric means, if the bath required 50 gal. of water raised in temperature by 50F. ? Solution. Cost = kw. hr., X $0.10 _ kw. sec. _ kw. sec, to raise 1 gal. 1 X 50 X 50 3600 3600 1 gal. weighs approx. 8.4 Ib. /. kw. sec. to raise 1 gal. 1 = 8.4 X 1.055 = 8.86 8.86 X 2500 ' ' kw ' hr ' = 3600 " = 6 ' 15 Cost = 6.15 X 0.10 = $0.615. Problem, 11. Four car heaters each take 4 amp. at 125 volts. Find the cost per 10-hr, day at lOc per kw. hr., to operate them on a 500-volt circuit, (a) when they are connected in series, (6) when they are connected in parallel. Answer. In series, $2.00; in parallel, $32.00. Problem 12. If the car contains 3000 cu. ft. and is insulated against loss of heat, how much time is required for a rise in temperature of 20C. when the heaters of problem 11 are connected (a) in series, (6) in parallel? Answer. In series, 12 min. 44 sec.; in parallel, 48 sec. NOTE. Specific heat of air at constant volume = 0.167. CHAPTER II FORM OF WORK In order that students may gain the greatest possible advantage from pursuing the course of study, it has been thought best to include in the body of the book, at this point, a brief statement of the procedure which the student should adopt in the working out of the problems. He is urged to familiarize himself with the method, and to follow it rigidly until, in so doing, he has thor- oughly acquired the habit of careful and accurate work. Object of Problems. Problems are almost universally con- sidered to be indispensable in any engineering course. Their function is similar in many respects to that of laboratory experi- ments. They illustrate the theory. In this respect problems may be divided into two groups, namely: (a) Those in which the general equation is applied to a definite concrete case, and (b) Those in which the general equation is investigated for the purpose of finding out the whole range of definite value which may be obtained from one variable by assigning definite values to one or more other variables. As an illustration of the first group, we will take the following example : Problem 13. Ten arc lamps, in series, are used to light a certain building. They require 6.6 amp., and the potential-difference (drop) across each lamp is 80 volts. Current is supplied from a power house 2000 ft. distant, by means of No. 6 B. & S. wire. If the energy is measured at the power house, find the cost at lOc. per kw. hr. to light the lamps 8 hr. per day. Solution. Cost per day = power X hr. X $0.10 Power, P = PL + Pw where PL power required by lamps = nEI where n = number of lamps and PW power lost in the wire Pw = I*R 9 10 ELECTRICAL ENGINEERING where R = total resistance of wire R = resistance per 1000 ft. X 1 -j~ resistance per 1000 ft. of No. 6 wire = 0.4 ohms at 75F. Then R = 0.4 X ~ = 1.6 ohms. PW = 6.6 2 X 1.6 = 70 watts PL = 10 X 80 X 6.6 = 5280 watts p = p L x P W = 5350 watts = 5.35 kw. Cost = 5.35 X 8 X 0.10 = $4.30. Ans. Such problems are typical of existing conditions. An engineer continually meets them where he is trying to find what results are being obtained from a given installation. In solving them, accuracy is the prime consideration, and this is obtained by avoid- ing short cuts and following through, step by step, a logical de- velopment. These problems are of far less importance and inter- est to the engineering student than problems of the second group. + 2000 r r H \ 1 ^VAAWWNAA^- t J> 1 1 250 ^50 Kw. %'/////'//ti FIG. 6. As an illustration of these take the following example: Problem 14. A load of 50 kw. at 250 volts is to be supplied by a power house distant 2000 ft. from the load. If the line costs 20c. per Ib. of copper laid, find and plot (a) efficiency of transmission against size of wire; (6) cost of copper against size of wire; (c) efficiency of transmission against cost of copper. load 50,000 load + line loss = 50 , 00 + /* 50,000 . . Efficiency = 50)000 X resistance per 1000 ft. = 4 X r X wt. per 1000 ft. = 4w. FORM OF WORK 11 Tabulation : Wire No. (B. & S.) 0000 00 1 4 8 12 16 r per 1000ft. 1 R = 4r 0.049 196 0.078 312 0.125 0.50 0.25 1.0 0.64 2.56 1.60 6.4 4.0 16.0 Wt. per 1000 ft = w wt. = 4.... '...;. Cost at $0.20 641 2,564 512.8 403 1,612 322 .4 253 1,012 202.4 126 504 100.8 50 200 40 20 80 16 7.9 31.6 6.32 40 00072 = 7 2 # 7,840 12,480 20,000 40,000 102,400 256,000 640,000 50,000 + 40,000/2 . . Efficiency 57,840 0.865 62,480 0.8 70,000 0.715 90,000 0.55 152,400 0.33 306,000 0.16 690,000 0.072 The curves are plotted in Fig. 7. 600 100 1 90 400 80 70 300 |60 ~ 1 i IB 50 J.I 200^40 30 100 20 10 OC < v ^ .-C09] 1 ^. \ \ ^ -^ >< X V 7 \ \ A I \ %! \ / \ ^. \ / N \ X \ [ ^ "^^. *" ^ 1 > 00 00 1 3 5 7 9 .11 13 15 17 Size of Wire ) 100 200 300 400 500 Cost. $ FIG. 7. Summary. The curves show (1) efficiency of transmission decreases as wire becomes smaller, at first slowly, then rapidly, and then, for very small wires, slowly again; (2) the cost of wire decreases as wire becomes smaller, at first very rapidly, then more and more slowly; (3) efficiency increases with cost, rapidly at first, for low efficiencies and costs, then more and more slowly. 1 From wire tables. 12 ELECTRICAL ENGINEERING It is evident that this problem could be greatly extended so as to include other variables, such as current density in the wire, cost of lost energy, etc., and indeed it is characteristic of this type of problem that there are always suggestive lines of investi- gation which tend to stimulate the student's interest. The work of solving the problem may be divided into a number of parts, thus: (1) statement of problem, (2) diagram of circuit, (3) analytical work, (4) tabulation of values, (5) plotting of curves, (6) summary, or statement in words, of the results obtained. The statement of the problem should be concise. The dia- gram should be an illustration of the statement, and should con- tain the symbols to be used. The analytical work should be carried out as far as possible with symbols before the numerical values are substituted. In the above example there is very little opportunity for the use of symbols, owing to the shortness of the problem. In later prob- lems this feature will be more apparent. Tabulation should be arranged with care, and should be planned so that columns can be conveniently added. As a rule, it is well to assign along the horizontal various values of the independent variable, and proceed, step by step, to the dependent variable. As in this case, there may be different combinations of variables, as number of wire, cost, and efficiency. This makes the tabula- tion more complex, as it would be by any other procedure, but it is still entirely clear. The plotting of curves is then carried out, and this should be done neatly and preferably in ink. The problem should then be completed with a brief statement of the results obtained. It is not always easy to make students take this last step, but they should be required to do so, and to follow this general plan throughout, until they have formed the habit of doing it and need no further compulsion. There may be other ways of working these problems efficiently, but it seems justifiable to urge teachers and students to adopt this method in preference to any other to which they are accustomed. It will insure uniformity and logical arrangement, will make cor- recting easy, and will commend itself to the student as well as the teacher. In working problems of this nature there are other objects than merely to illustrate and enforce the theory. Great stress is laid on them, not only for the engineering knowl- FORM OF WORK 13 edge which they contain, but because of their structure, which, it is believed, strongly tends to develop those qualities most essen- tial in an engineer. For instance, the mathematical develop- ment calls for insight and understanding, the tabulation calls for concentration of mind, the summation of results calls for accuracy, and a study of the plotted curves calls for judgment. At the same time, efficiency, the keynote of the engineer, would be lacking if the problems were not done in the shortest and best way consistent with obtaining the desired results, and it is obvious that many hours will be wasted, both to student and instructor, unless the work is done with order, accuracy and neatness. CHAPTER III MAGNETISM FARADAY explained magnetic phenomena by assuming that surrounding a magnet or a wire carrying current were lines of force. The stronger the magnet or current, the stronger is the magnetic field, that is, the more lines of force per square centimeter. The introduction, then, of a magnet into a space means the establishing of a field of force. To get quantitative ideas about field strength he made use of the symbol H which was called the intensity of the field, or the force on unit pole placed in the field. It seems an unfortunate term since intensity and density are readily confused. B, the density of the field, or the number of lines of force per square centimeter, is proportional to H, and also to a quantity n, the permeability or magnetic conductivity of the medium in which the intensity, H, exists. Thus B = fj.H. = 1, it follows that the number of lines of force per square centimeter is numerically the same as the intensity of the magnetic field H. H, the force per unit pole, is expressed in dynes. The force exerted on a pole not of unit strength, but of Jr IG. o. . strength m, is F = mH dynes, where, of course, H is caused by other poles than m. Consider, now, an isolated elementary pole of strength m, from which n lines of force, per unit pole, protrude radially and uni- formly in all directions (Fig. 8). 14 MAGNETISM 15 At a distance r from ra, no matter what the medium is provided it is uniform, the density of the field is B T %, since the area of a sphere of radius r is 47rr 2 . The force, H, on unit pole is then _, , , . jr, Thus the force on pole mi is r B nm COULOMB, working in air, found experimentally that the force between two poles of strength m and mi could be expressed by p = k y-, and he would have found F = k ^ had he experi- mented in a medium of permeability, /*. Therefore, k may be written unity if n = 4ir. In other words, if it is assumed, as is the case, that 4?r lines protrude from unit pole. GAUSS came to the same conclusion from another point of view, and the relation = 4irm is called GAUSS'S theorem. In words, GAUSS'S theorem states that from a pole of strength m radiate outward 47rm lines of force, or the total outward flux, $, from pole m is 4irm lines. 1 Cylindrical Poles. To find the intensity of the magnetic field H at a point distant r from a uniform cylindrical pole of strength m (Fig. 9). By GAUSS'S theorem the, flux = 47rm, and # = The area of a cylinder of radius r and length I is , . _ 2m at p, is _ flux area* 2irrl. Thus 2m , = -> and \ 1 t I FIG. 9. FIG. 10. Flat Poles. To find the intensity of the field, H , at a point dis- tant d from one side of a flat pole of strength m (Fig. 10). As- sume that the lines of force are perpendicular to the surface. Let B = flux density at any distance. Then the flux coming from one of the surfaces of the magnet is < = ^ = 2irm. 4 If the area of the pole face is S. then B = -> and H = FT- S MO 1 For a more complete discussion of GAUSS'S theorem see "Advanced Course in Electrical Engineering." 16 ELECTRICAL ENGINEERING Magnets as Commonly Used in Meters. To find the magnetic intensity between poles (Fig. 11). Let S be the area of a pole face and d the distance between poles. The density in the gap between the two pole faces is due to the magnetic north pole, N, as well as to the south pole, S. If the lines of force flow outward from the north pole, they flow inward from the south pole. Thus a simple examination will show that the fluxes add in the gap and cancel each other in the outside region. The total flux from N is 4irm and one-half of this flux is assumed to be in the gap, the other half extending outward. The density in the gap due to N will then be _ 4?rm _ 2-rrm ^ n " : ~2S~ ~S~' Similarly, due to S, and = # = 4?rm M Bfj. In all practical problems where magnets act in air only, /* is, of course, unity. Consider, now, the pull between the faces of a magnet as shown in Fig. 11. The flux density at the south pole due to the flux from the north . . n 2irm 2irm , . , pole is B n = g . . H = -g- = force on unit pole at the sur- face of the south pole. Since the south pole has a strength m, the force on it is therefore ' do) Usually the density, B, in the gap is known. Substituting the value of m from (9) into (10) gives 27r ~ 4V or the force in dynes per sq. cm. is F = ^ MAGNETISM 17 In air, where ju = 1, B 2 F = - dynes per sq. cm. (11) In Ib. per sq. in., the formula becomes It is seen that if the pole strength, m, remains the same while the faces of the magnet approach each other, the density, and thus the force, is constant. The work done is then Fd, where d is the distance between the TT/ B * Sd poles, or W = ~ Energy Density in a Field. The volume of space through which the body is moved is Sd. The energy density, or joules per cu. cm. of space between poles, is then : 1 B 2 B z * 3 = S ergs = 8000* J0ules ' The conception of energy density is merely mentioned at this point. Similarity of magnetic and eledtric fields will be shown later on together with the development of theory and problems in electro-statics. Limits of Pole Intensity. In practice it is found that the limits to which pole intensity -~- can be pushed are as given in the fol- lowing table: TABLE II. APPROXIMATE LIMITING VALUES OF o For wrought iron magnets, 1600 units of pole strength per cm. For soft steel magnets, 1600 units of pole strength per cm. For cobalt magnets, 1300 units of pole strength per cm. For nickel magnets, 500 units of pole strength per cm. For permanent steel magnets, 800 units of pole strength per cm. The Magnetic Cycle. According to the molecular theory of magnetism, magnetic bodies are composed of minute magnets which attract and repel each other, and which are partly free to turn under the influence of magnetizing forces. When strongly magnetized, these molecular magnets are pointed in the direction of the magnetic force. When the force is removed, they still tend to point in the same direction, and thus the body exhibits magnetization, which is called residual magnetism. 18 ELECTRICAL ENGINEERING The magnetic state of a body is shown with reference to the "magnetizing force" by a curve called the hysteresis loop (Fig. 12). Magnetization of an iron bar is ordinarily accomplished by sending current through a number of turns of wire wound around the bar. The magnetization is thus produced by the ampere- turns (A.T.). The number of lines of flux set up per unit area enclosed by the turns will with a long bar be shown to be r / ' M = B, where ju is the permeability of the bar and I is its length. Since in air /-t = 1 and H = B, it follows that the intensity of the magnetic field in a solenoid is: QAirA.T. The hysteresis loop is drawn with flux density, B (in lines per square centimeter or per square inch), as ordinates and the mag- netic field intensity, H (or frequently, for convenience, ampere- TT7\ turns per inch length of magnetic circuit, j J , as abscissae. The construction of the loop is as follows: Imagine a bar of iron wound with many turns of insulated wire. If the iron has no residual magnetism at the beginning, be- fore current is sent through the wire, there will be no magnetizing force and no flux, and consequently the first or starting point on the curve will be at a (Fig. 12). As more and more current is sent through the wire, that is, as the magnetizing force is increased pro- portionally to the current, the flux or induc- tion density, B, is increased, not according to a simple law, but in such a way as to give the characteristic curve (1) from a to 6. If the magnetomotive force (m.m.f.) expressed in ampere- turns is now decreased, the curve (1) is not retraced, but B follows curve (2) from b to c. At c, H = 0, while B continues to have a value represented by the line ac. This value of B corresponds to the residual magnetism of the iron. If, now, the current be reversed, so that H is given negative values, B continues to decrease from c to d. At the point d, B = 0, while H has the negative value ad. This value of H is MAGNETISM 19 called the "coercive force" of the magnet. It is the magnetizing force necessary to reduce the remanent magnetism, ac, to zero. As H is further increased, negatively, B follows the curve de. At 6, which corresponds to b with positive H, the current is again reduced, and B follows curve (3) to /, which gives the value, of, of negative remanent magnetism corresponding to ac for H = 0. Thus, the point a is not reached again, but as H is now given increasing positive values, the curve goes through g to b, complet- ing the loop. In obtaining a single loop, the points do not usually come into such close agreement, due primarily to the fact that there is always some remanent magnetism at starting, which prevents the curve from beginning exactly at a. But in the case of many uniform reversals of H, as occurs in electrical machinery, the loop is retraced uniformly so long as the limiting values of H re- main constant. It will be later shown, in connection with the study of hysteresis losses, that the area enclosed by the loop is proportional to the work done on the magnet per cycle. T> Permeability. The ratio -p is called the permeability, and is a measure of ease with which lines of flux are set up in a given material. Permeability is denoted by the symbol /*. Numer- ically, B = H in air (or vacuum) since /* = 1. In the magnetic metals, particularly iron, steel, nickel and cobalt, ju undergoes wide variation in value, with different values of H . For a more complete discussion of the subject of magnetism the student is referred particularly to EWINQ'S "Magnetic Induction in Iron and Other Metals." CHAPTER IV N PRINCIPLE OF THE ELECTRIC MOTOR A wire carrying a current was discovered by OERSTED to be surrounded by a magnetic field, which is strongest near the wire. A small needle, placed in the field (Fig. 13), is directed along the lines of force, but there is practically no tendency for it to move toward the wire as the forces of attraction exerted on its poles are equal and opposite. A long needle, however, tends to move toward the wire as there is a component of force on each pole in the direction of the wire. A wire carrying current, placed in a field perpendicular to the lines of force (Fig. 14), causes the flux to be distorted, and this tends to force the wire in such a direction that the lines shall again take up their normal position. This is the principle of the electric motor. The electric motor consists (Fig. 15) of a number of wires wound on a drum, and so placed in a magnetic field that the current is caused to flow downward (toward the plane of the paper) on, say, all the wires adjacent to the north pole, 1 and up- ward on all the wires adjacent to the south pole. The wires on FIG. 13. N N FIG. 14. FIG. 15. the left, then, tend to move downward, and those on the right upward, and thus rotation is produced. 1 In the diagram a cross, <8>, is used to represent down-flowing current and a dot, O, up-flowing current in accordance with notation in common use. 20 PRINCIPLE OF THE ELECTRIC MOTOR 21 The current which, when flowing in a wire 1 cm. long placed at right angles to a field having a density of 1 line per sq. cm., gives a force of 1 dyne is called the abampere. The force, in dynes, is then F = IIB where / is the current in abamperes, I the length of wire in centi- meters, and B the flux density of the field in lines per square centi- meter. The force is due to the interaction of flux and current. If, however, the lines are not at right angles to the wire, B must be replaced by its component which is at right angles to the wire. If the angle is a (Fig. 16), then the force is F = IIB sin a, where B sin a is the component of flux at right angles to the wire. Problem 16. A copper wire carrying 10 amp. is placed in a magnetic field of 10,000 lines per sq. cm. What is the force in pounds on each centimeter of the wire (a) if it lies perpendicular to the direction of the magnetic field, (6) if it lies parallel to the field, (c) if it makes an angle, a, with the direction of the field? N FIG. 16. (a) Solution. F = IIB sin a F, per cm. = IB sin a. Sin a = sin 90 = 1 / = 10 amp. = lab amp. B = 10,000 .*. F, per cm. = 10,000 dynes. 10,000 dynes = == 10.2 grams 10 0.02245 Ib. 453.6 (6) Sin a = sin = .'. F, per cm. = 0. (c) For any angle, a, F, per cm. = 10,000 sin a dynes = 0.02245 sin a Ib. Determinations of Magnetic Intensity. Magnetic intensity at the center of a coil (annulus) . Let a magnet pole, m, be placed at the center of a coil (Fig. 17).' It will send out lines in all directions, some of which will strike an element of the coil, dl, 22 ELECTRICAL ENGINEERING at right angles. Thus a force, dF, will be generated in the direc- tion of the axis, as indicated, and its value will be dF = IBdl = / dl snce The total force on the coil will be - n dl 2irrlm This will be the force, due to m, with which the coil will tend to move along its own axis. It is obviously also the force on m due to the coil. Thus if a unit pole (m = 1) replaces the pole of strength m the magnetic field intensity at the center of the coil is found. It is: a _*a*,*L r 2 r Magnetic Intensity at Any Point along of a Coil. The force dF will act angles to the line joining m and h m ~ * F ji---" I* 10. 18. dF = IBdl = Idl d dF has components, dF cos a and dF sin a where sin a = and a sin The component of force which tends to move the coil in the direction of its axis is dF sin . Call this component dFi. Then and For Jp m . / I. sin 3 .adZ = 27rm I sin 3 m = i p l = H sin 3 a PRINCIPLE OF THE ELECTRIC MOTOR 23 since the component dF cos a is balanced around the coil and thus exerts no force. Magnetic Intensity in the Center of a Long Coil. The force at m, due to an element of the coil, dx (Fig. 19), is dF = 2irml sin 3 a , where / is the current in abamperes, in the ele- ment dx. FIG. 19. If the current per centimeter length of the coil is / c , then , ,._ 2irml c dx sin 3 a I = I c dx, and dF = x / 1 \ But = cot a. Differentiating, dx = r ( =-^ ) da. Substituting this value of dx in the formula, 2irml c r sin 3 a , dF = ^ da r sm 2 a = 2irm7 c sin ada. Co. - ai .*. F = I 27rm/ c sin ada = 4irl, ./a =" TT ai m cos or H = 47r/ c cos ai. For very long coils, cos ai = 1, and H = 47T/ C . The relation between H and the ampere-turns of a coil may be found as follows: Let there be a current of I abamp. in the coil, and let n = number of turns. Then nl = abamp.-turns. Abamp.-turns nl per cm. = -r- = J c , where I = length of coil in centimeters. 47m/ 1 I Then H = 4ir7 c When the current is in amperes, 1 NOTE. It should be noted that the above value of H holds only for infinitely long solenoids since it was derived on that assumption. For practical purposes, according to the accuracy required, this value of H may be 24 ELECTRICAL ENGINEERING H j , whence, amp.-turns = 0. SHI (12) i If I is in inches, amp.-turns = Q.313HL When the coil has an air core, H is numerically equal to B, and amp.-turns = Q.SBl or = 0.313J5Z. 1 FIG. 20. FIG. 21. Application of Magnetic Formulae to Instruments. Let a rec- tangular coil of height, a, and width, 6, be suspended in a magnetic field of uniform density, B (Fig. 20). The two sides, a, are per- pendicular to the flux, and therefore, with a current of / abamp., there will be a force on each wire of F a = IBl = IB a dynes. FIG. 22. This force will produce a torque around the axis, on each wire, of T = IBa ~ dyne-cm. The total torque per turn = 2T a = I Bab, i and if there are n turns, T = InBdb, dyne-cm. In practical instruments, T should be about 1 gram-cm. Let a circular coil of radius r, as in Fig. 21, be suspended in the field. To find the torque on any element dl. The useful part used whenever it is desired to find the magnetic field intensity along the axis of a solenoid, and not very near the ends, provided the length of the coil is about 50 times its diameter. It is necessary to observe this (always depending on the accuracy desired) on account of the disturbing effects of the ends, as can be easily seen by comparing the figures. (Fig. 22.) 1 NOTE. It can be proven that the density in the middle of such long coil is uniform, thus if A is the area inside of the solenoid the total flux is AB, area PRINCIPLE OF THE ELECTRIC MOTOR 25 of dl is its component perpendicular to the lines of flux, = dl sin 6. Then, dF = InB sin Bdl. This force acts with a lever arm = r sin 6, and the torque is therefore dT = InB sin 6 X r sin 0dl. But dl = rd0. Hence dT = InBr* sin 2 0d0, and T = I 7nr 2 sin 2 Jo [01 T 27r - - 7 sin 20 Zl 4 J = InBr*ir = InB A y where A = area of the loop. In practice, permanent magnets are generally used to produce the flux. CHAPTER V DESIGN OF A LIFTING MAGNET It has been shown that for a path of magnetic lines in air, the following relation obtains: amp.-turns = 0.313 Bl", if inch measurements are used. For an iron path, the necessary ampere- turns are obtained from a curve of B vs. AT, where B is the flux density. Such a curve called either magnetization or " satu- ration " curve, is obtained experimentally from a sample of any desired magnetic material. The curve thus obtained will be approximately correct for that material, but variations are always 100 120 Ampere-Turns per Inch FIG. 23. juo 160 180 liable to occur due to either physical or chemical influences by which any portion of the material is made to differ from the sample used to derive the curve. By testing many samples a typical curve is obtained for any given material. In Fig. 23 is given a set of these satura- tion curves of iron and steel as commonly employed in electrical machinery. 26 DESIGN OF A LIFTING MAGNET 27 Let it now be required to make the calculations for the design of a cast-iron electromagnet to lift a weight of 1000 Ib. through a gap of 1.5 in. Let it be assumed that the magnet core is of the shape and dimensions given in Fig. 24. Then the area of a pole face is 10 X' 5 = 50 sq. in. The I two pole faces have an area of 2 X 50 = 100 sq. in. Then the weight to be lifted per square inch of area is enclosed in the loop. Thus j ," - "'''.-. e= ~ d it ' Consider now that the loop revolves in a uniform magnetic field. When the loop encloses the entire field it may be said to be in the zero position. Let it be assumed that in zero posi- tion it encloses 100 lines. In position 1, displaced 10, it will then enclose 98.5 lines. The loss of 1.5 lines from the loop has resulted in a generated e.m.f. or, in general, dt t% ti At 29 30 ELECTRICAL ENGINEERING If the coil rotates at the rate of 1 r.p.s., the time required for it to move 10 is : of 1 sec. = 0.0277 sec. = A. Then e = 1*5 0.0277 " 0.0277 = 54a6 volts. This procedure may be followed and a tabulation made for every 10, so as to obtain data from which to plot, point by point, an e.m.f. wave, thus: Angular position of coil 10 20 30, etc. Flux enclosed, < 100 98.5 94.0 86 6 Change of flux, z , in the amount of flux enclosed by the coil, per unit length of the coil parallel to the shaft. d = 2Bdx is the change in flux per centimeter length of coil, due to both con- ductors. Then d = 2Brd0 sin 0, since ds = rd0, and dx ds sin 0. Substituting, kr sin 0d0 k sin 0d0 nax = FIG. 28. 4r 2r sin 4 - 2 sin Then, _ d __ JL^ /2k sin 0d0 \ k s e " "" dt ~ dt U - 2 sin 07 " 2T^ fc sin sn Sin may be written sin 2irnt, where is expressed in radians and 2irn denotes angular velocity. Then -j=r = 27rn, and n is in revolutions per second, or is frequency jn a two-pole machine. For machines of any number of poles -jj = 2wf where / = frequency. k sin 2 - sin X2wf. Let Then and 100 volts. This occurs when = ~ 100 = Substituting this value of k, _ e max 2-n-f sin _ e max sin = 27r/ X 2 - sin " 2 - sin ' Problem 18. Calculate and plot the e.m.f. for th$ above condition, for one-half wave. E.m.f. Wave when the Coil is Wound on an Iron Core. In all these cases it is sufficiently correct to consider only the lengths 32 ELECTRICAL ENGINEERING of the flux path in the air. By following the general procedure of the preceding paragraph, the e.m.f. of a coil on the iron core is found to be (m 1) sin where m - sin 2mr m = 7^-' or D Problem 19. Calculate and plot for one-half wave, the e.m.f. for this case, when e max = 100 and m = 1.1. (Fig. 29.) Additional Problems for the Determination of E.m.f. Waves. It is very good experience for the student to work out and plot a number of these waves. For this purpose a few additional problems are suggested. Problem 20. Determine the e.m.f. of a coil wound on a wooden drum when B max = 100 lines per sq. cm., speed = 1 r.p.s. and the dimensions of the dynamo are as given in Fig. 30. Dimensions are given in centimeters. Plot the wave, point by point, for each millimeter of distance across the pole face. _L Problem 21. On an iron armature between rectangular poles as in Fig. 31, let two coils, at right angles to each other (that is, in space quadrature), be joined in series, so that their e.m.f. waves add. Plot the resultant e.m.f. per centimeter length of the armature. Show that the resultant e.m.f. due to the two coils is less than their sum. Problem 22. Same as last, but for three coils spaced 120 apart. Problem 23. Continue the development of the method of the last two problems and finally obtain the average value of the e.m.f. of an armature whose conductors are spaced uniformly around the periphery. CHAPTER VII FIG. 32. INDUCTANCE Inductance. When a circuit connected to a source of e.m.f., e t is closed through a switch, S (Fig. 32), a current is established in the coil, and sets up a magnetic flux which links with the turns of the coil. This flux produces a back, or counter e.m.f. in each turn, e = -TJ-, or in the N turns, e = N - Expressed in volts this is N_ d$ 10 8 dt Inductance is defined as the number of iijterlinkages of flux with turns, per unit current, or, in symbols, Expressed in practical units it is: e = (13) L = where i is the current in amperes. Problem 24. Let a coil of 200 turns be supplied with various amounts of current, and let the flux produced, when 1 amp. flows, be 1000 lines. Find the inductance. Tabulating from Eq. (14): i i 2 4 10 tf). . 1000 2000 4000 10000 N N 200 0.2 X 10 200 0.4 X 10 6 200 0.8 X 10 6 200 2 X 10 fl T 0.2 X 106 0.002 0.2 X 10 6 0.002 0.2 X 10 6 0.002 0.2 X 10 0.002 1 Students almost invariably have difficulties with inductance. This problem is given solely for the purpose of impressing upon them the fact that inductance is a "constant" of the circuit. 3 33 34 ELECTRICAL ENGINEERING From the values of I/, thus obtained, it is seen that L is a constant, and is independent of the current; it is a " circuit con- stant" similar to resistance in a coil having a non-magnetic core. Transposing Eq. (14), "J&- Differentiating with respect to time, di _ N d L Jt"l0 8 ~dt Substituting into (13), which is the common expression for induced electromotive force, or counter e.m.f. of self-induction. 1 In Fig. 33 is represented a circuit of resistance, r, and inductance, L, which is oj connected to a source of e.m.f., e. When 1 *- o *' the switch, S, is closed, the e.m.f. has Tfl _ qq to overcome the resistance, r, and also the counter e.m.f. of self-induction due to the setting up of flux in the coil. Therefore we may write: e = ir + L^ (15) This equation is fundamental, and is general for circuits possessing only resistance and inductance of constant value. An algebraic relation between the impressed e.m.f. and the current assuming in this case that the e.m.f. is kept constant is found as follows: e-ir 1 In an ironclad magnetic circuit the inductance is not a constant. It depends upon the permeability. The flux is not proportional to the current producing it but is a complicated function thereof. In that case ^ = ~ (Li) = L ~ ' + i^. INDUCTANCE 35 /. t = - ? log (e - ir) + C, whence rt log (e - ir) = - + Ci, and " + Cl - r ^ e ir L = C 2 L (16) C 2 is determined from the nature of the problem. The rate of energy supply or power equation corresponding to (15) is obviously ei = i z r -f Li r \ Jo and the energy supplied by the generator is T eidt. The energy dissipated in heat is f*T \ i*rdt Jo and the energy supplied and thus stored in the magnetic field is JT j S*T Li^di= ( Lidi = y 2 LP where I is the value of the current at time T. Starting and Stopping Current in an Inductive Circuit. Re- ferring to equation (16) it is evident that for t = 0, i = when starting the current and i = I for t = in stopping the current, since energy cannot be altered in an infinitely short time and therefore current cannot be established or changed in an infinitely short time. Thus when considering the starting of a current we have for t = 0, i = 0. Substituting these values in (16) = - r [e - C 2 e] = -i [e - Cj, 36 whence, and ELECTRICAL ENGINEERING (17) This equation gives the value of the current at any instant after the closing of the switch, S. If the impressed e.m.f., e, is sud- denly short-circuited by the closing of the switch, S' (Fig. 34), then e = 0, and, at the instant of closing, t = 0, FIG. 34. * = I, where Z is the current in the circuit just before closing the switch. Substituting these values into (16) whence, and 7 = [0 - CJ, C, = - rZ, (18) This equation gives the current at any instant, t, as it is dying away in the circuit after the e.m.f. has been suddenly removed. The inductance of coils varies with the size, shape and number of turns. If a given length of wire of defi- nite size is to be made into a coil, maximum inductance F IG< 35. will very nearly be obtained if the coil has the proportions given in Fig. 35. 1 the inductance, in this case, will be: 0.27 cm. 2 The value of 10Xc , henrys > where the length of the coil is given in centimeters, and c is in centimeters. 1 BROOKS and TURNER, "Inductance of Coils," Bulletin No. 53, Univ. of Illinois Engineering Experiment Station. INDUCTANCE 37 Problem 26. Find and plot current vs. time when the circuit is closed on a coil of 1 km. of No. 15 B. and S. wire (diam., d.c.c., = 0.066 in.; r/1000' = 3.17 W ), designed for maximum inductance, e = 100 volts. Problem 26. Find and plot the curve of dying away of the current when the coil of problem 25 is short-circuited. Problem 27. Find the average value of the inductance of the lifting magnet previously designed (Chap. V), and determine how long it will take for the current to rise to 90 per cent, of its permanent value. CHAPTER VIII ALTERNATING CURRENTS It has been shown that the fundamental equation in an induc- tive circuit where the resistance and inductance are constants and not depending upon the current is : This equation gives the relation between the particular values of e.m.f. and current at any instant. In the case previously discussed it was assumed that the im- pressed e.m.f., 6, was constant. In most engineering problems the e.m.f. is, however, not con- stant but it varies from instant to instant. Almost all electrical instal- lations now use alternating current rather than direct current. In this case it will be seen that the e.m.f. and current can almost always be as- FIG. 36. j- L i sumed to vary according to a simple sine law. In other words it can be assumed that the instantaneous value of the current at any time, t, can be found from equation i = I m sin ut (Fig. 36), where o> = 27r/ = angular velocity, and / = frequency of alternation of the current = number of cycles, or complete reversals, per second. I m = maximum value of cur- rent. For 60 cycles, o> = 27r60 = 377. .'. i = I m sin 377*. Differentiating eqv: i = I m sin w we get di - = 7 m w cos wt. Substituting in (15), e = r I m sin wt + L 7 m w cos orf = I m (r sin wt + Lw cos o>0 (19) 38 ALTERNATING CURRENTS 39 Thus e is the sum of two component waves, one depending on the sine of ut, and the other on the cosine. Problem 28. Let I m ponent waves of e.m.f. 1, r = 0.5, Leo =0.4. Find and plot the com- Sin at I m r sin ut = ir Cos wt . . w cos L* dt e. 00 00 00 40 40 30 .5 .25 .866 .346 .596 60 0.866 0.4330 0.5 0.2 0.6330 90 1.0 .5 0.0 0.0 .5 120 0.866 0.433 -0.5 -0.2 0.233 150 0.5 0.25 -0.866 -0.346 -0.096 180 0.0 0.0 -1.0 -0.4 -0.4 These waves are shown plotted and combined in Fig. 37. Problem 29. A similar set of waves should be obtained by each student from values of I mi r, Leo, assigned at random. By inspecting these waves it is seen that i lags behind e, that FIG. 37. is, it passes through zero later than e by about 40. This il- lustrates one of the characteristic features of inductive circuits. It should also be noted that ir is in time phase with i, and that iLu is in time quadrature with i, being 90 in phase ahead of i. The quantity Lo> is called reactance. It is measured in ohms, and denoted by the letter X. Thus Lw = X, where L is the in- ductance in henrys, w = 2ir/ is the angular velocity in radians per second, and X is the reactance in ohms. X is not, like L, a property of a coil or circuit, but depends on the frequency. The average value of the e.m.f. generated in a coil of a dynamo, depends only on the speed of rotation and the number of lines of 40 ELECTRICAL ENGINEERING flux cut; that is, it depends on the average rate of cutting of the lines of flux, by the conductors, and not on the distribution of the lines under the poles. The effective value of e.m.f . does, how- ever, depend on the distribution of the flux. Frequency has been defined as number of cycles per second. A two-pole generator, at 1 r.p.s. has the frequency,/ = 1. A four-pole generator, at 1 r.p.s., has/ = 2. fry A p-pole generator, at N, r.p.s., has/ = -^N. The coil, in position 1 (Fig. 38a), contains the whole flux. The coil, in position 2, contains no flux. Thus, a change of the whole flux takes place in a quar- ter of a revolution. If T is the time of 1 cycle, the whole flux is therefore cut in the T time r - 4 The average rate of cutting is then <|> 4$ FIG. 38. ~m" ~7p where & is the total flux. T Therefore, the average e.m.f. is -y-, where N is the number of turns, and 2N is the number of conductors per circuit. At 60 cycles, In general, a-.-L * 60 .'. Average e.m.f. = 4N3>f = 4N$f X 10~ 8 volts. In a four-pole machine (Fig. 386) all flux is cut in % revolution. The average rate of cutting is therefore ~, w here I i T\ is the time of a revolution. Average rate = -^ = 83>N g = ~^- = 4*/ where N s is the number of revolutions per second, and f = - N s . With N turns, average rate = 4/A r = average e.m.f. X 10~ 8 volts (20) ALTERNATING CURRENTS 41 This equation is identical with that for a 2 pole machine. It applies regardless of the number of poles as long as N is the number of turns in series per circuit. Average Value of a Sine Wave. The e.m.f. induced by rota- tion of the armature conductors in the field is Let

$ TO sin ut is the rate of change of the flux, and e = volts. sin ut. In practical units, sin 10 8 Since co = 2irf this may be written, sin co/ 10' For maximum e.m.f., sin u* = 1, and E m = ^ s ' m volts (21) To obtain the average value of e.m.f., integrate a half-wave and divide by TT, that is, by the length of a half-wave. Thus, . = - ( "sin 6dd = -\ - cos 0| = - = 0.636. irjo TT|_ Jo 7T 2 2 .'. The average value = - X E m . Multiplying (21) by - 7T 7T Av. e = - which agrees with the average value previously found (20). Effective Value of a Sine Wave. Let i = I m sin 6. If this current flows through a resistance r, it has been seen that the heat developed at any instant is i 2 r. Thus, the heat developed per cycle may be expressed as, T 2 % r I I 2 m sin 2 Ode. Jo By trigonometry, sin 2 = y>, V<> cos 20. 42 ELECTRICAL ENGINEERING Substituting, The average value of energy flow or the rate at which energy is being dissipated, or the power, is rlmir = rim 27T 2 r/i Thus, the rate of heat dissipation is -- The effective value of the current corresponds to a constant or direct current which would give the same heat in the same time if flowing through the same resistance. . = , and i ett . = = 0.707 I m . Similarly, the effective value of e.m.f. is obtained, and e e ff. = 0.707 E m , where E m is the maximum value of the sine wave of electromotive force. effective value . , . , - I The ratio - is called the form N average value factor. j With sine waves form factor (j(f) = QQ^Q = ^_ . . The equation for the effective value of the I e.m.f. is obtained from (21) by multiplication i Fro. 39. ' m . m - '"~ V210 ^0^^ This applies to a concentrated coil of N turns. If, however, the turns are distributed over the periphery, as in a direct-current armature, from Fig. 39 it is seen that coil 1 contains all the flux, while coil 2 contains the flux X cos 6. Therefore the .effectiveness of coil 2 is N c cos 0, where N c number of turns of the coil. Let N = total number of turns. Then the turns per cm. of N armature periphery = ^ , where r = radius of armature. ALTERNATING CURRENTS 43 N Then the effectiveness of the turns per cm. is ~ cos 6. The average effectiveness of the turns per cm. is NcosB.. N r+i 1 I N cot ,J l.r N 2 The total effectiveness is therefore 2rr X -^ = -N. Therefore, in a distributed winding, the turns are not so effective as when they are concentrated. Thus, for distributed winding, 4 44fAT **m ^ CHAPTER IX DIRECT -CURRENT GENERATORS Homopokr Generators. These are also called by the names " acyclic" and "unipolar." They are a small class of machines, distinguished from the usual types of direct-current machinery in that the conductors always move through the magnetic field in a constant direction with respect to the direction of the lines of flux. Among the earliest of dynamos may be mentioned one of this type known as " FARADAY'S Disc Dynamo," in which a copper disc was rotated between the poles of a permanent magnet. Current was collected by means of two brushes making contact, respectively, with the rim and axle of the disc (Fig. 40) . A more modern type of homopolar generator is shown diagrammatically in Fig. 41. For the permanent mag- FIG. 40. I, \ ; i ! f \ < i N E 3 r p 1 F ^ - 1 Q f L_ ^ , > ^ ) ^ V ( f. ~""\ J \ _s FIG. 41. net is substituted a powerful electromagnet, and two sets of brushes are used instead of one. By connecting these brushes in series outside of the machine the total e.m.f. at the terminals is doubled. 44 DIRECT-CURRENT GENERATORS 45 From the fundamental considerations developed in Chap. VI it is evident that the voltage between each set of brushes is e 10 8 where N = revolutions per second and 3> = total flux, since there is only one conductor between the brushes. With any ar- rangement which permits the use of additional sets of brushes, as in Fig. 41, the voltage is increased in proportion to the number of sets of brushes connected in series, and becomes e = 10 8 FIG. 42. where c is the number of conductors and is equal to the number of sets of brushes in series. Fig. 42 illustrates the use of bar conductors on the armature. Each conductor is connected to two slip rings on which brushes bear. There are thus twice as many slip rings as conductors. Since the conductors are in- sulated, they may be put in series by properly connecting their brushes outside of the machine. Direct-current Machines with Commutators. On most direct-current machines use is made of commutators. To understand these machines a knowledge of the principle of wind- ings on the armature is needed. In Fig. 43 a single coil is repre- sented in a magnetic field. The ends of the coil are connected to the segments of a two-part commutator. In the position 46 ELECTRICAL ENGINEERING shown, the e.m.f. is maximum. As the coil moves in the field, the segments move under the brushes and the e.m.f. at the brushes, AB, during a half revolution, has the values of a half- sine wave. When this e.m.f. reaches zero, the segments pass from under the brushes. The same operation is then repeated and gives a succession of half waves, all in the same direction. If now another loop is placed on the armature at 90 to the first one, a new series of half waves will be added at 90 to the first series. By connecting these loops in series, suitably joining to commuta- tor segments and continuing to use only two brushes, the e.m.fs. of the loops are added together and produce a resultant e.m.f. shown in heavy dots by the wave "d" (Fig. 44). This wave never reaches zero and is much more steady than that produced by a single coil. By continuing this process, all irregularities are virtually wiped out and there results a smooth wave of constant e.m.f. A simple example of armature winding with commutator and brushes FIG. 44. is shown in Fig. 45, for the purpose of illustrating the connec- tion of coils in series. Types of Direct-current Commutator Machines. Direct- current machines are usually divided into groups according to the method of exciting the field magnets, as follows: 1. Permanent Magnet Machines. These have no field windings, but the field structure consists of hard-steel permanent magnets. They constitute a small group, used chiefly for telephone sig- nalling and gas engine ignition. DIRECT-CURRENT GENERATORS 47 2. Separately Excited Machines. In these, the field winding is supplied with current from an external source. The chief advantage of this type is that it enables a steady field excitation to be maintained at all times regardless of the fluctuations in voltage at the brushes. 3. Shunt Machines. In this type the source of excitation of the field is derived from the terminals of the machine itself. The field circuit is connected in parallel with the external cir- cuit and the field current varies as the voltage of the machine changes. 4. Series Machines. The current in the armature is made to flow also through the field windings; that is, the field and armature coils are connected in series with the external circuit. Thus the field excitation is proportional to the load current. 1 Permanent Magnet 2 Separate Excitation 3 Series FIG. 46. 4 Siunt 5 Compound 5. Compound Machines. These are excited partly by a shunt winding and partly by a series winding, each pole being provided with both a shunt and series coil. The total field excitation thus depends upon the voltage of the machine as well as on the load current. These five types are illustrated in Fig. 46. Other com- binations are sometimes used in special cases. The performance characteristics of these various types of generators differ greatly. In general, the characteristic of a generator is a curve showing the relation between terminal voltage and the load current, the latter being the independent variable. These curves and others of a similar nature should be thoroughly studied, especially in the laboratory. Armature Reaction. When a generator is delivering no cur- rent the direction of the field flux is along the axis of the poles. 48 ELECTRICAL ENGINEERING When current is flowing, however, the armature becomes an electromagnet on its own account, and the field flux becomes the resultant of that produced by the field windings and that due to the armature winding. Fig. 47a shows a bipolar dynamo with a ring armature. Arrows show the direction of current and also of flux. Starting from the negative brush, the current divides as it enters the armature, one half winding around to the left, the other half pursuing a similar path to the right, and both finally joining again to enter the positive brush. It is to be noted that the flux set up by these armature currents is, in general, in space quadrature to the flux due to the field winding. FIG. 47. Fig. 476 shows an equivalent diagram representing a drum armature. When once the principles of current action in the armature are understood, it is simpler to make use of the rep- resentation of Fig. 476 than of Fig. 47a. For clearness, the com- mutator is omitted in the case of the drum, the position of the brushes being indicated with reference to the armature itself. It makes no difference how the end connections are made, so far as the armature m.m.f. is concerned. In this case, since the brushes are not shifted but are placed on the so-called neutral axis midway between the poles, the armature magnetomotive force is directed vertically upward, while the field magnetomo- tive force is, as always, along the pole axis. The resultant magnetomotive force is the vector sum of these two. Since the armature m.m.f. acts at right angles to the field m.m.f., its effect is said to be wholly cross-magnetizing. When the brushes are shifted a the armature m.m.f., still acting along the brush axis, may be resolved into components. DIRECT-CURRENT GENERATORS 49 p c = p A cos a, the cross-magnetizing component, acting at right angles to the field, and FD = FA sin a, the demagnetizing component, acting directly in opposition to the field. The re- sultant m.m.f., OR, Fig. 48, is then due to the m.m.fs. of the field OF and the armature OA, the latter being composed of OC, cross-magnetizing, and OD, demagnetizing. Cross-magnetization is always present when the armature carries current. It distorts the field and displaces the neutral axis, necessitating thereby a shifting of the brushes. When the brushes are shifted, demagnetization also enters in, weakening directly the field strength. Under such conditions the resultant flux takes up a general direction as indicated by the shading in the air gap in the figure. The pole tips are unequally magnet- ized, the leading tips being weakened and the trailing tips strengthened. A c FIG. 48. The actual direction of the resultant flux is not along OR but along OR' ', a line of somewhat less deviation from OF. This is because of the unequal reluctances of the paths along the direc- tions of the component m.m.fs. Consider, for example, a generator whose flux per pole enter- ing the armature is < r , under conditions of normal operation, that is, voltage and speed. To generate this flux at no load would require F amp. -turns on the field core if all the flux generated in the field passed through the armature. Some flux, however, passes around the armature without cutting its conductors. This is called leakage flux, and amounts to 15 or 20 per cent, of the net flux, in ordinary machines. To provide this leakage flux as well as the net flux, where p = number of poles. CI e The total armature amp.-turns per pole = -~~' Let the brushes be on the geometrical neutral, that is, midway between the poles. Then, since the conductors are distributed over the entire periphery, the effective armature A.T. per pole 9 r T r = / L = ffk. TT c 2p irp The effect of these distorting ampere-turns has been shown to be to weaken the flux in the leading pole tips and to strengthen that in the trailing tips. The net result owing to unequal saturation of the iron, is to reduce the actual amount of the flux. In order to compensate for this reduction extra ampere-turns must be placed upon the field core to the amount of about 40 per cent, of the armature cross-magnetizing ampere-turns. Thus, F c = > if there is no brushshift irp and the total field amp.-turns per pole are F t = kF. + -^. vp When the brushes are shifted a, the cross-magnetizing amp.- turns are F c = ~ X -j~ (Fig. 49). . 180 - 2 I C C Their effective value is k c r^ TT~ loU 4p where r* Ur 2 cos a cos dO, = 7T where = IT 2a The demagnetizing turns consist of a belt of conductors of width 2a. The effective demagnetizing amp.-turns are then , 2a I C C DIRECT-C URRENT GENERA TORS 51 where /+! 2 sin a cos ede = ~2^- These latter ampere-turns act in direct opposition to the field. If there were no leakage of flux between field and armature, they would be compensated by placing an equal number of additional ampere-turns on the field. Owing to leakage this number must be multiplied by k. The total required field ampere-turns under the condition of brush shift of a and I c amp. in the armature conductors is then (90 - a)I c C / ( 0.4 in order that the flux entering the armature shall be

overcome armature r jrns eaction 10,427 Armature Reaction. " Armature reaction" means effective ampere-turns per pole on the armature. The actual amp.- turns per pole, in this case, are 167 X 36 = 6000. Since the turns are distributed over the armature surface the effective amp.-turns are - X 6000 = 3820. 7T DESIGN OF A DIRECT-CURRENT GENERATOR 61 If there were no shift to the brushes, these ampere-turns would all be cross-magnetizing, or distorting. To compensate for them, it is necessary to supply about 40 per cent, of their value in additional ampere-turns on the field core. It is assumed, however, that the brushes will be shifted 15, giving a distorting belt of 180 - 30 = 150. To overcome the distorting ampere-turns at full-load there will then be required 240 200 .120 40 7 2000 4000 6000 8000 Ampere-Turns FIG. 52. 10000 12000 150 180 X 36 X 167 X k e X 0.4 = 1480, where + 2 k c = ~ \ cos ed0 = x^ [1.9318] = 0.737. The demagnetizing ampere-turns constitute a belt 30 wide. To compensate for them would require their exact numerical equivalent, were there perfect mutual induction between these turns and the field. Owing to magnetic leakage there should be added about 15 per cent, to the effective demagnetizing ampere- 62 ELECTRICAL ENGINEERING turns. To compensate for these, therefore, will require 30 X 36 X 167 X fed X 1.15 = 1140 amp.-turns, where k d = 0.99. To overcome armature reaction at full-load will require 1480 + 1140 = 2620 additional amp.-turns on the field core. For any other load, keeping the same shift, the required ampere-turns will be proportional to the load current. No-load and full-load saturation curves are shown in Fig. 52. The Shunt Field Winding. Under no-load conditions it is evident that the shunt field current must supply the entire ex- citation. In this machine, therefore, the shunt field m.m.f. must consist of 7500 amp.-turns per pole when an e.m.f. of 250 volts is being generated. Actually, each shunt spool is wound with 460 turns of No. 7 B. & S., D.C.C. wire. The field current is therefore = 16.3 amp. The shunt coil has an actual length of 6.25 in. As the di- ameter of No. 7 wire is 0.16 in., including insulation, there will be Q-TQ X 6.25 = 39 turns per layer of wire. There will be -^ = 11.8 layers, or practically 12 layers, giving a depth of winding of 0.16 X 12 = 1.92 in. The mean radius of the coil, allowing for spool thickness, is then Mean radius = - ' ~ - = 5.46 in. .*. Mean length of turn = 2ir X 5.46 = 34.35 in. Total length of wire on each shunt spool is -r^- in. X 460 = 1316 ft. Resistance of No. 7 wire at 65C. = 0.586 ohm per 1000 ft. .'. Resistance of each shunt spool is 0.586 X 1.316 = 0.77 ohm. The resistance of the entire shunt field is r f = 0.77 X 12 = 9.24 ohm. DESIGN OF A DIRECT-CURRENT GENERATOR 63 The voltage drop on the shunt field is i/r f = 16.3 X 9.24 = 151 volts. The voltage drop in the shunt field rheostat is e rh . = 250 - 151 = 99 volts. The Series Field Winding. Consider two cases: (1) the genera- tor to be flat-compounded, (that is, the no-load and the full- load voltages are equal, as specified), and (2) the generator to be 5 per cent, over-compounded. In the first case, it is evident that the shunt field ampere- turns will remain the same at full-load as at no-load since the same voltage, 250, is impressed on the shunt circuit. But by Tables III and IV, it is seen that at full-load there will be required 10,427 7500 = 2927 additional amp. -turns. These must evidently be supplied by the series field m.m.f. FIG. 53. The actual winding consists of 2>^ turns per pole. Each turn is made up of 5 strips of conductor in parallel, each strip being 3^ in. wide by 0.095 in. thick. The accomplishment of half a turn is illustrated in Fig. 53 which represents the arrangement, in plan, of the series field winding. The series field current must then be 2927A.2 7 . l * = ocx " = H70 amp. 2.5 turns This means that with full-load current 2000 - 1170 = 830 amp. must be diverted from the series turns by a shunt con- nected in parallel with them. This shunt is, in practice, usually composed of German silver strips whose length is so adjusted by test as to divert exactly the required amount of current. In the second case, the full-load voltage with 5 per cent, over- compounding is 1.05 X 250 = 262.5. To obtain this voltage requires the addition of 11,700 7500 = 4200 amp.-turns to the no-load ampere-turns. This additional 64 ELECTRICAL ENGINEERING excitation is not all supplied by the series field m.m.f., however, since the shunt field current is affected by the increased terminal voltage. The shunt field m.m.f. now consists of 1.05 X 7500 = 7875 amp.-turns. Therefore, the series field m.m.f. must consist of 11,700 - 7875 = 3825 amp.-turns. The current in the series winding is then i t = o g = 1530 amp. a.O The current diverted through the shunt to the series field is 2000 - 1530 = 470 amp. The shunt field current is if = 17.1 amp. Consideration of the saturation curves will show that this is nearly the limit of over-compounding for this machine. If full-load voltage of 275 were desired, it would be necessary to add another half turn to each series coil. The series field m.m.f. has been made to compensate for the armature reaction and the ir drop (assumed 2% per cent.) in the armature. So far as the field design is concerned, this is satisfactory. These calculations are, however, only approxi- mate and the actual values should now be determined from the known data of the machine. Armature Resistance. Being multiple wound, there are 12 paths in parallel in the armature. Each path includes 72 con- ductors, or 36 turns. The length of a turn is twice the gross length of the armature plus the end connections. The end connections for one turn may be taken as 9 X diameter per pole of the armature = 9 X 5.33 = 48 in. Length of one turn is thus 2 X 9 in. + 48 in. = 66 in. f\f\ \/ Q \ Length of one path of 36 turns = - - = 198 ft. iZi Since the area of each effective conductor section is 0.0675 sq. in., its resistance is found to be 0.142 ohm per 1000 ft. at 65C. Resistance of one path is thus 0.142 X 0.198 = 0.02812 ohm. Resistance of 12 paths in parallel is " = 0.00234 ohm. DESIGN OF A DIRECT-CURRENT GENERATOR 65 The true armature resistance will be somewhat less than this owing to the intermittent short-circuiting of coils by the brushes, and its average value may be taken as r a = 0.00226 ohm. Voltage drop in the armature is e a = r a l a = 0.00226 X 2017 = 4.55 volts. Brush Resistance. There is always a drop in voltage at the brushes due to the true brush resistance and also to the re- sistance of the sliding contact between brushes and commutator. This combined resistance has no definite value which may be calculated, but it is found by experiment that the drop which it causes amounts to 2 volts when the current density in the brushes is 30 amp. per sq. in. or more, while for densities less than 30, the drop is proportional to the current density. 30 amp. per sq. in. is about the usual current density in brushes. Drop across brushes is thus 6b = 2 volts. Series Field Resistance. Total thickness of series conductor = 0.095 in. X 5 strips = 0.475 -in. Area of series conductor = 0.475 X 3.125 = 1.485 sq. in. Mean radius of series turn, allowing ^32 m - insulation between turns, is found to be 5.12 in. mean radius. 3 turns + mean radius, 2 turns Mean radius = - 2 (4.5 + 0.475 + 0.0313 + 0.233) + (4.5 + 0.475 + 0.0156) 2 5.24 + 5 R1 _. 2 = 5.12 m. .*. Mean length of series turn = 2 X 5.12 X TT = 32.2 in. ,, , . . ,. 12 X 32.2 X 2.5 Length of series winding = r~ - = 80.5 ft. approx. To this should be added about 5 ft. for connections between coils, making the series winding 85.5 ft. long. Resistance per 1000 ft. of series conductor is found to be 0.00645 ohm at 65C. Series field circuit resistance is therefore r a = 0.00645 X 0.0855 = 0.000552 ohm. As it was found that only 1170 amp. go through the series field coils at full-load, the voltage drop on the series field wind- ing is e a = r 8 i a = 0.00055 X 1170 = 0.645 volt. 66 ELECTRICAL ENGINEERING Total voltage drop in the machine is therefore e a + e b + e 8 = 4.45 + 2 + 0.645 = 7.095 volts. or ' = 0.0284, or, approximately, 2.5 per cent, as assumed. /oU If the assumption of percentage drop is not considered to have been sufficiently close, the magnetic calculations should be repeated using the new percentage just found. Commutator and Brushes. The size of the commutator is determined chiefly by the brush requirements. The number of commutator segments is 432, that is, one segment to each effective turn on the armature. The brushes rest perpendicularly on the commutator. There are 12 studs of brushes, each stud holding 10 brushes. Each brush has a cross-section of 1.25 in. X 0.75 in., giving a brush area of 0.94 sq. in., or 9.4 sq. in. per stud. As there are six positive and six negative studs, the area of the positive (or negative) brushes is 6 X 9.4 = 56.3 sq. in. Therefore the current density in the brushes at full-load is 2016.3 , a = 35.8 amp. per sq. in. OD.o The commutator length must exceed that of the brushes on the stud, that is, it must exceed 10 X 1.25 + some space of separation between adjacent brushes. In this case the com- mutator length is 17.5 in. The commutator diameter is influenced by the peripheral speed. Being built up of numerous copper segments each separated by sheets of mica, the commutator is usually mechan- ically weaker than any other revolving part. It must not only be protected from forces which would cause it to fly apart, but there must be no force acting upon it which will be strong enough to cause even slight warping of its surface. Good com- mutation demands smooth, even contact between the segments and the brushes at all times. On the other hand, too small a diameter results in very narrow segments, thin and wide brushes and then, in turn, a longer commutator. The commutator diameter for this machine is 39 in., which is approximately 60 per cent, of the armature diameter. From DESIGN OF A DIRECT-CURRENT GENERATOR 67 this it is found that the width of segment plus the mica in- sulation is 7r39 432 = 0.284 in. 0.75 The brushes will therefore extend over Q * = 2.64 segments. Flux Distribution Around the Armature. It is of interest at this point to investigate the distribution of the flux around the armature periphery on account of its bearing on the commuta- tion and also in order to be able to determine the potential difference between any two adjacent commutator segments. This is best accomplished with the help of a diagram in which is shown a pair of poles drawn to scale in relation to the armature, developed along the horizontal line. FIG. 54. A curve abcde, Fig. 54, is first constructed to represent the flux distribution around 360 electrical space degrees of the armature periphery. This curve is based on the assumption of flux density, being inversely proportional to the flux path in the air. Thus, the density is uniform under the pole and is so represented by the line ab. To determine the densities between the poles, empirical mean flux paths to the teeth are drawn, and the flux along each path is taken as inversely proportional to its length. The curve cde will obviously be the reverse of curve abc. 68 ELECTRICAL ENGINEERING The second step is the construction of a curve of armature magnetomotive force. This m.m.f . will act in the direction of an axis midway between the poles (assuming brushes to be set on the geometrical neutral). Along this axis the m.m.f. will consist of all the armature ampere-turns per pole. Acting through the next adjacent teeth s, s, the m.m.f. will be diminished by the amount of armature ampere-turns included between these teeth. These ampere-turns may be plotted, tooth by tooth, in the manner thus indicated, and the result will be a curve, fgh, in the form of successive steps corresponding to the armature teeth. To construct the flux curve of the armature reaction from the m.m.f. curve, reluctance of the air paths alone need be considered. To be sure, the rest of the flux path, especially that of the teeth, would have some effect on the accuracy of the curves so obtained. But the error would not be great, being anywhere from 2 per cent, to 8 per cent, according to the position of the point on the curve. The flux density for each tooth is therefore determined from the formula: 3.19A.7 7 . B - - z , where I is the length of the path in air. 1 This is plotted as curve, ijk, to the same scale as the curve of the field flux density abode. The actual densities along the periphery will vary from tooth to slot, and, indeed, this variation is noticeable on many oscillo- grams of alternator voltage. The ripples which occur in the flux wave due to alternate teeth and slots would exist equally with reference to the field flux, armature flux and resultant flux. In order t6 avoid confusion the ripples have not been shown on the armature density curve, but all the waves are plotted as smooth lines. A study of the resultant wave reveals the great distortion caused by the armature current, the strengthening of the flux in the pole tips at A, the weakening at B, and the shifting of the neutral point, c, in the direction of rotation. The student may well discuss fhe effect on the flux density waves of giving a shift of the brushes. Losses and Efficiency. The .efficiency of a machine is given by the equation, output efficiency = , = ---- DESIGN OF A DIRECT-CURRENT GENERATOR 69 The full-load output of the generator has been given P = El -*- 1000 = 500 kw. as The problem of the efficiency is then one of determining the The losses of a generator may be considered under three heads: (1) copper losses, due to heat developed by the currents in the windings; (2) core losses, due to hysteresis and eddy cur- rents set up by the changes of magnetic flux in the iron and, to a slight extent, in the copper of the machine and (3) friction losses, including that of the bearings, the brushes and windage. Copper Losses. These consist of I 2 r loss in the armature, the shunt field circuit including the rheostat, the series field coils, and that of the brushes and commutator. It is not sufficient to ascertain these losses for full-load only. The quality of a generator is displayed by its performance at all reasonable loads. The efficiency will in this case, therefore, be calculated for loads from zero to 150 per cent, of full-load. The armature copper loss is 7 a V a , where r a = 0.00226 ohm. Shunt field copper loss is I/Ef, where E/ is the voltage impressed on the field circuit, and is in this case 250 volts. // = 16.3 amp. .'. Shunt field copper loss = 16.3 X 250 = 4075 watts. The series field loss is I S E S , where I 8 is the line current = I a If, and E a has been found to be 0.645 volts at full-load and varies directly with // for other loads. TABULATION OF COPPER LOSSES Per cent. 25 50 75 100 125 150 load la 16 516 1,016 1,516 2,016 2,516 3,016 /a 2 256 266,000 1,037,000 2,300,000 4,075,000 6,330,000 9,100,000 / 2 r 0.58 600 2,340 5,200 9,200 14,300 20,550 ItEf 4,075 4075 4,075 4,075 4,075 4,075 4,075 L 500 1,000 1,500 2,000 2,500 3,000 E, 0.161 0.322 0.484 0.645 0.806 0.968 I.E, 81 323 725 1,290 2,018 2,905 E b 0.6 1.2 1.8 2 2 2 I a E b 310 1,220 2,730 4,032 5,032 6,032 Total loss 4,075 5,066 7,958 12,730 18,597 25,425 33,562 70 ELECTRICAL ENGINEERING Brush loss = I a Eb, where Eb is the voltage drop in com- mutator and brushes, being approximately proportional to cur- rent density in the brushes up to a density of 30 amp. per sq. in. and being 2 volts for higher current densities. Core Loss. The hysteresis loss is principally in the armature and is due to the reversal of direction of the flux in the metal as the armature spins around. The amount of energy expended in reversals of the magnetic molecules is proportional to the frequency and approximately proportional to (flux density) 1 - 6 . Thus, Hysteresis loss = kfB 1 - 6 . The exponent 1.6 was found experimentally, by STEINMETZ; it holds with sufficient accuracy for the usual range of flux densities obtained in electrical machinery. In direct-current armatures hysteresis loss usually amounts to about 2.8 watts per Ib. at/ = 60 and B = 64,500. Assuming this value as standard, the armature core loss and teeth loss are ex- pressed by the equation W h = 2.8 X j4 X (54 KQQ) L8 X wt. of core or teeth in Ib. For both core and teeth, / = 37.5. B, in core = 43,600 at 250 volts, full-load. B, corresponding to average amp.-turns required by the teeth = 126,000. Weight of armature core = vol. X wt. of 1 cu. in. = 0.28 X 6.05 X 7r(30J 2 - 22 2 ) = 2430 Ib. Weight of teeth = 0.28 X 6.05 X k(32 2 - 3O7 2 ) - 216 X 1.3 X 0.465] = 0.28 X 6.05 X [258 - 130] = 217 Ib. Substituting these values, the total hysteresis loss in teeth and core is = 1.75 [0.676 1 -* X 2430 + L95 1 - 6 X 217] = 3340 watts. The eddy current loss is due to the heating of the core by local or eddy currents set up in the material of the core by the chang- ing flux within it. DESIGN OF A DIRECT-CURRENT GENERATOR 71 It is therefore an I 2 R loss, or -~-, where E is the e.m.f. set up, which is expressed by the equation where = B X area = total flux. From this it may be seen that the eddy current loss may be written We = k^B*. The eddy current loss may be reduced as much as desired by making the laminations of the armature core sufficiently thin. A satisfactory value for this loss may be obtained by assuming it equal to the hysteresis loss. In that case, W e = 3340 watts and the total core loss is W c = 2 X 3340 = 6680 watts. Losses in pole faces and copper due to eddy currents are here too small to consider. There will also be slight changes in the values of the core loss as the load changes, due to variation in magnetic densities, especially in the teeth. This variation is also slight, however, and will be neglected. Friction Losses. Loss due to brush friction is based on a coefficient of friction of 0.3, and a brush pressure of 1.2 Ib. per sq. in. of brush surface. From this, the friction per sq. in. is 0.3 X 1.2 = 0.36 Ib. Surface area of one brush = 1.25 X 0.75 = 0.9375 sq. in. Total brush friction force is then, F = 0.9375 X 10 X 12 X 0.36 = 40.5 Ib. Power loss, h w " - Hm x 746 watts - where r = radius of commutator in ft. = 1.625 and n = speed in r.p.m. = 375. Thus, _ 2w X 1.625 X 375 X 40.5 X 746 Wb = 33,000 = 35 WattS ' 1 This is the equation for induced e.m.f. in a transformer. It holds also in this case as will appear later when the transformer is studied. 72 ELECTRICAL ENGINEERING Bearing friction and windage, together, make a complicated loss to determine with accuracy. This loss is, however, one which may be assumed with quite sufficient accuracy from the data obtained in practice. A fair assumption to make for generators of this type is 1 per cent, of the rated output of the generator. In this case then W f = 500,000 X 0.01 = 5000 watts. Summary of Losses, Output and Efficiency. The combined losses of the generator for different per cent, loads is given in Table V. '60 20 50 75 % Load FIG. 55. 100 125 150 TABLE V The efficiency curve is shown plotted against per cent, load in Fig. 55 Per cent, load 25 50 75 100 125 150 Copper loss 4,076 5066 7 958 12 730 18 597 25 425 33 562 Core loss 6680 6 680 6 680 6 680 6 680 6 680 6 680 Friction loss .... Total loss 8,500 19,256 8,500 20,246 8,500 23 138 8,500 27 910 8,500 33 777 8,500 40 605 8,500 48 742 Output Input 19,256 125,000 145,246 250,000 273,138 375,000 402,910 500,000 533,777 625,000 665 605 750,000 798 742 Efficiency 86 915 93 936 939 939 Temperature Rise. The final limit to the output of the generator is the permissible temperature rise. The effect of DESIGN OF A DIRECT-CURRENT GENERATOR 73 temperature on copper is to increase its resistance to a slight extent; the effect on iron is to increase its permeability. These effects tend to offset each other so that, as tar as these two materials go, it would be permissible to attain very high temperatures. . On the other hand, the insulation is the real limiting feature. Of the many insulating materials, none possesses the com- bination of qualities necessary in the ideal insulator for electrical machinery. This material should be of high insulation strength, strong mechanically, and its insulating and mechanical qualities should not change under long-continued heating. Mica is the best insulator in these respects, except that it is poor from the mechanical standpoint. Asbestos is useful owing to its heat-resisting qualities, but it is a rather poor insulator and its mechanical possibilities are limited. Cotton tapes .and varnishes do not withstand the high temperatures. In attempting to extend the limit of output of machines of a given size there are two lines along which lie the main pos- sibilities of success. Either some new insulating material, more satisfactory than those at present in use, may be discovered or invented, or im- provement in ventilation and heat radiation may be accom- plished by alteration of the mechanical design. Under existing conditions a temperature rise of 40C. above that of the surrounding air is quite conservative. The tem- perature which different parts of a machine will attain is hard to predetermine accurately from the design. Practical studies have afforded certain empirical constants which permit ap- proximate determinations to be made, but in any case, practical experience will greatly assist the designer in his attempts to keep close to the limits. For the present it will be sufficient to determine the watts per square inch of surface of field spools and armature. For rotating machinery 0.5 watt per sq. in. will correspond roughly to a temperature rise of 40C. The external surface of a field spool, only, should be taken, and the same applies to the armature. These should, of course, be calculated separately. Problem 30. In the machine just studied, show by calculation, as indi- cated above, that the temperature rise in the field and armature coils will not be excessive. CHAPTER XI ELECTRICAL CONSTANTS OF A DIRECT-CURRENT GENERATOR HAVING COMMUTATING POLES AND COMPENSATING WINDING As a typical generator of this more complex type will be taken the following: M.P. 6 - 1000 - 600 - 1200/1260 volts. The generator is thus 5 per cent, over-compounded. Being designed for comparatively high voltage, commutation becomes a matter of special importance. To insure proper neutralization of the armature reaction, there- fore, special field windings are supplied, and these are so placed as to counteract the armature m.m.f. in space as well as in amount. That is, neutralization is accomplished by means of a compensating winding placed in the pole faces symmetrically with respect to the armature conductors under the pole arc, and an auxiliary commutating pole inserted between the main poles, where the armature magnetomotive force is the strongest, and whose duty is not only to neutralize this magnetomotive force about the neutral point in which the brushes are placed, but to supply a flux which will be in proper direction to balance the e.m.f. of self-induction of the commutated coil. With such an arrangement the brushes are given no shift, and, consequently, the armature m.m.f. is entirely cross-magnetizing. The series field m.m.f. proper is thus relieved of every duty ex- cept those of compensating for IR drop in the armature and over- compounding. The circuit diagram of this machine is given in Fig. 56. General dimensions and specifications are as follows: Armature outside diameter, 48 in. Armature inside diameter, 28 in. Armature gross length diameter 15.5 in. Armature effective diameter, 11.7 in. Armature ventilating ducts, 4% in. wide. 2% in. wide. 74 ELECTRICAL CONSTANTS 75 Slots, number and dimensions, 144; 0.44 in. X 1.53 in. Effective conductors per slot, 6. Effective armature conductor section, 0.55 in. X 0.09 in. Armature winding, multiple drum. Compensating Commutating Series Winding Poles Rheostat Compensating Winding 1 FIG. 56. Yoke section, rounded, 17 in. X 6.5 in. Main pole core section, 14.5 in. X 14.5 in. Main pole core length, including pole shoe, 14 in. Main pole core length, allowed for field spool, 13 in. Main pole arc, 17.5 in. 76 ELECTRICAL ENGINEERING Commutating pole section, 13.5 in. X 2.25 in. Commutating pole length, 14 in. Main air gap length, 0.3125 in. Air gap under Commutating pole, 0.5 in. Shunt field winding; 2256 turns per spool of No. 15 B. & S. triple cotton-covered wire. Series field winding; 3 turns per spool. Each conductor built up of 4 strips giving total section, 1.5 in. X 0.35 in. Commutating pole winding; 5.5 turns of copper ribbon 12 in. wide X 0.05 in. thick. 1400 1000 800 600 400 200 400 600 800 Ampere-Turns FIG. 57. 1000 1200 Compensating (pole face) winding consists of 16 conductors per pole contained in 8 holes in the pole face. Each hole has 2 conductors, one, a tube, the other a rod within the tube. Tube outside diameter, 1% in., inside diameter 2 % 2 in. Rod diameter, % in. Commutator diameter, 30 in. , Commutator length, 14 in. Commutator segments, 432. Segment width, 0.219 in. ELECTRICAL CONSTANTS 77 Brushes per stud, 7. Brush section dimensions, 1.25 in. X 0.875 in. The armature flux at no-load is readily found to be 13.9 megalines per pole. The no-load saturation curve is given in Fig. 57, having been determined in exactly the same manner as that of the previous machine, given in Fig. 52. This curve shows that 7600 amp.-turns are required to give normal voltage at no-load. At full-load, the shunt field m.m.f. 1260 will supply J2QO X 760 = 798 am P-- turns - Assuming 2 per cent, voltage drop in armature and brushes, the total e.m.f. which must be generated is 1.02 X 1260 = 1285 volts. From the saturation curve, this voltage requires 8750 amp.-turns. Therefore the series field m.m.f. must supply 8750-7980 = 770 net amp.-turns per pole. To supply these, however, account must be taken of the un- fortunate situation of the series field winding with respect to magnetic leakage. Being placed close to the yoke, the leakage factor should probably be 1.50 instead of 1.25 as used for the shunt field calculation. This factor could, of course, be calculated, but it is hardly desirable to introduce such a refinement when the means of adjustment of the series field current render a reason- able assumption entirely satisfactory. On the basis of a leakage 1 50 factor of 1.50, the series amp.-turns are T^ X 770 = 924. 924 The series field current is I 8 = -$- = 308 amp. o Current diverted around the series field is I d = 793 - 308 = 485 amp. The entire load current of 793 amp. passes through the 9 turns per pole of the compensating winding, and the 5J^ turns of each commutating pole. SATURATION CURVE CALCULATION No-load. E = 1200. _ E x 1Q8 *" = 4/< where 600 6 * = 60 X 2 = 30 ' 78 ELECTRICAL ENGINEERING 6 X 144 t = x 2 = 72 turns per pole. 1 200 ^ 1 = 13 > 880 > 000 = flux in teeth 307 X 72 X 4 and gap at no-load. The flux in the pole core is fa = 1.25 X 13,880,000 = 17,340,000, where 1.25 is the leakage factor. It is fairly large in this case, as is usual when interpoles are present. Tooth width at face = Teeth per pole = Teeth width (face) Teeth width (base) Teeth area (face) Teeth area (base) Gap area Arm. core area X48 144 - 0.44 = 1.05 - 0.44 = 0.61 in. pole pitch 10.85 in. X 10.85 = 10.15 in. 24 X X 1.08 25.5 = 17.75. 17.75 X 0.61 in. 44.92 48 10.85 X 11.7 = 127 sq. in. 10.15 X 11.7 = 118.8 sq. in. 3 X 17.5 X 14.5 + 127 = 222 sq. in. 44.92 - 28 X 11.7 = 99 sq. in. Pole core area = 14.5 X 14.5 = 210 sq. in. Yoke area = 17.5 X 6.5 X 0.95 = 108 sq. in. No-load. E = 1200 Part Mate- rial Flux Area B AT fin. Length AT Teeth (face) Teeth (base) .... Gap.. 13.88 13.88 13.88 127 118.8 222 109,200 116,800 62,500 93 \ iso/ 136 19,600 1.53 0.3125 208 6,125 Arm Sheet Pole iron . . Steel 6.94 17.34 99 210 70,000 85,200 7 33.5 9.5 14 67 470 Yoke Steel 8 67 108 80,300 30 24.3 730 Total 7,600 E = 600 Teeth (face) 54,600 3.9 1 . Teeth (base) . . . 58,400 4.4 I 4 ' 15 6.35 Gao 3,063 Arm . . . 35,000 2.4 22.8 Pole 41,250 9 126 Yoke 40,150 8.8 215 Total 3,433 ELECTRICAL CONSTANTS E = 900 79 Teeth (face) .... 82,000 11 Ol Teeth (base) . . . 87,500 14 5/ 12 ' 75 19.5 Gap.. 4,594 Arm . . 52,500 3 77 35 8 Pole 61 900 15 9 222 5 Yoke 60,200 15.1 367 Total 5 239 E = 1400 Teeth (face). 127,400 420 1 Teeth (base) 136,100 1,264 f 842 1,290 Gap. 7,150 Arm 81,600 10.8 102.7 Pole 96,200 79 1,107 Yoke .... 93,600 63.8 1,660 Total 11,310 FIG. 58. Calculations of armature, shunt and series field windings, as well as brush losses and friction loss are made in exactly the same manner as in the preceding example. The difference in location of the shunt and series windings is given in Fig. 58. The division of the shunt into two coils per pole is made to 80 ELECTRICAL ENGINEERING allow the necessary room for end-connections of the compensating winding. The calculation of the commutating pole winding is likewise a matter of applying the old principle. The conductor itself is of extreme dimensions, being a band of sheet copper 1 ft. in width. For the compensating winding the mean length of 1 turn is found to be 2 X (length of pole parallel to shaft + 4 in. (extension)) + 2 X mean span between poles, = 2 X (14.5 + 4) + 2 X 19 in. = 75 in. Total length of winding = -^ X 8 X 6 = 300 ft. Area of conductor section = 0.442 sq. in. Resistance of winding = p- = -TQJ- X TTTIo = 0-0054 ohm. Voltage drop in winding = 793 X 0.0054 = 4.28 volts. Loss in winding = 4.28 X 793 = 3400 watts. Voltage drops and losses at full-load in other parts of the generator are as follows: Voltage drop Loss Armature ........... ........ ..... 14.75 11,700 Shunt field ........................ (1,008) including rheostat ................ ...... 4,500 Series field ........................ 0.636 505 Compensating winding .............. 4 . 28 3,400 Commutating field ................. 1.19 945 Brushes (7 2 #) ..................... 2 1,590 Brushes (friction) .................. ...... 1,760 Hysteresis loss ..................... ....... 7,220 Eddy current ...................... 7,220 Friction and windage ............... ...... 10,000 Total voltage drop ............. 22 . 856 Per cent, voltage drop ...... .... 1 . 82 Total energy loss .... r ......... 48,840 output 1,000,000 Efficiency = = 1,048,840 = Efficiencies for all loads are as follows: Per cent, load 25 50 75 100 125 150 Percent, eff. 88.5 93.5 94.85 95.3 95.5 95.5 Fig. 59 shows the efficiency curve. ELECTRICAL CONSTANTS 81 60 40 25 50 75 100 % Load FIG. 59. 125 150 O Q O Ok /O 000 FIG. 60. 82 ELECTRICAL ENGINEERING EFFECT OF COMPENSATING WINDING AND COMMUTATING POLES To study the effect of these windings in neutralizing armature reaction, it is best to construct a curve of magnetomotive forces showing their distribution along the armature periphery. From this and the curve of field flux density the resultant flux density along the periphery is obtained. Such curves are given in Fig. 60. The armature ampere-turns and field flux density in the gap are plotted to separate scales as was done in Fig. 54. The corn- mutating pole and compensating winding ampere-turns are like- wise plotted, but their direction is, of course, opposite to that of the armature m.m.f. The resultant m.m.f. of these three is given by the heavy irregular line. The average of this resultant m.m.f. is seen to be very nearly zero, showing the effective compensation of the armature reaction. It is also observable that the commu- tating pole m.m.f. is made sufficiently strong to overbalance con- siderably the armature m.m.f. in the neutral axis, thus creating a resultant flux oppositely directed to the armature m.m.f. The maximum armature m.m.f. which acts along the commu- tating pole axis is 9564 amp.-turns. Opposing this is the m.m.f. of the compensating winding which is 6344 amp.-turns, and the m.m.f. of the commutating pole which is 4360 amp.-turns. Thus the resultant amp.-turns amount to (6344 + 4360) 9564 = 1140. When the armature is in the less advantageous position (that is, with a slot in the commutating pole axis), the resultant amp.-turns are 10,704 - (11.5 X 797) = 1554. The average resultant amp.-turns along the commutating pole axis are therefore 1350. These ampere-turns, acting through a gap of % m - produce a flux density of 1350 B = 0.313 X 0.5 = 862 lines per Sq " in ' Commutation. If the field in the neutral axis were completely neutralized, commutation would still be poor due to the reversal of the current in the conductors during the period of commuta- tion. Therefore, to balance the e.m.f. induced in the short- circuited coil under the brush, an approximately equal e.m.f. is created in the opposite direction in this coil by causing it to cut through the flux due to the commutating pole. Exact neutralization of the induced e.m.f. in the short-circuited ELECTRICAL CONSTANTS 83 coil is practically impossible by this means. Current in the conductors does not vary logarithmically as in an ordinary circuit when the impressed e.m.f. is removed. If it did, the fundamental equation, (15), e = ir + L ^, where r and L are approximately constant, would apply for the induced e.m.f. of the coil. But in this case, r is by no means constant due to the varying brush surface on the commutator segments. The value of r is therefore some func- tion of the time. Putting r = / (Q, and considering the varia- bility of L, due to change of permeability in the iron part of the flux path, the induced e.m.f. would be expressed by the equation e = 7(0 + I (Li)- To solve this equation to a satisfactory degree of approxima- tion, certain assumptions may be made. First, let it be assumed that the current dies down in the coil as a sine wave (Fig. 61). The in- duced e.m.f. would then be maximum when the coil axis passed through the center of the brush. If this maximum value were de- termined, it could be made equal to F the e.m.f. produced by rotation of the coil through the field set up by the commutating pole. Other values than the maximum could be left to care for themselves, being of secondary importance. The maximum value of the e.m.f. of self-induction is where I is the current in the armature conductor at the moment when commutation begins, and, in this case, is 133 amp. and X is the reactance of the coil. The second assumption is that L, the coil inductance, is constant. Hence X = 27r/ c L, where f c = frequency of current during commutation. f c = ^TTT where T c is the time of cummutation, since this time "I c evidently corresponds to one-half wave length. The time of 84 ELECTRICAL ENGINEERING commutation is that time taken by the commutator to move a distance equal to the thickness of a brush. In the machine under consideration each brush covers four segments. /. T c = -- X segments covered by brush r.p.s. total segments 1 4 = TO X 432 = ' 000952 sec ' and 2 X 0.000925 = 54 cycles per sec " In calculating the coil inductance, L, it is not sufficient to consider only the interlinkage of each coil with the flux which it produces. Mutual induction is also present, the value of L desired being therefore not strictly the self-inductance, but including that due to the interlinkage of the flux produced by the current in all 6 turns with each single turn. In this machine the conductors in each slot are all in parallel; thus N is 1 turn, composed of 2 conductors. It should be noted that of the 2 conductors com- posing any turn, one of them lies in the lower half of its slot, while the other lies in the upper half of its slot. The interlinkage of each conductor with the total flux will not be the same in the 2 cases. However, by considering the total flux as due to the 67 amp.- turns of a slot acting through an effective magnetic conductance, G, and surrounding each of the 6 conductors, the inductance thus calculated will be correct, provided the proper value of G is determined. Thus, = 67 X G, where < is the equivalent flux surrounding all conductors in 1 slot. (The inductance due to end-connections must also be ascertained, as is done later.) The magnetic conductance per centimeter effective length of the armature is calculated by means of the general formula, area x Considering the magnetic circuit (Fig. 62), it is seen to consist of 3 parallel paths in air, namely: that of section A and length B through the conductors, that of section C and length B above ELECTRICAL CONSTANTS 85 the conductors, and that of section F and length D, from the top of 1 tooth to the top of the other. The common path through the iron may be neglected as offering comparatively little resistance. To find the effective magnetic conductance per centimeter length across the section A , con- sider an elementary section, dx, at distance, x, from the bottom of the conductors. The con- ductance across this section is ' QAirdx B FIG. 62. The amp. -turns acting in this conductance are T-, where / is ** in amperes. Therefore the flux set up through dx is d QAirdx ~B~ X 2.47T/ x dx AB ' This flux interlinks with only -r- conductors. A. Nd + = - L -AJ8 ' 14.47T/ r\ , z 2 cfo Thus, and 14.47T/A SB is the interlinkages of the flux with all 6 conductors. Thus, if the flux is considered to be due to 67 amp.-turns acting through conductance, g, and this interlinks with each conductor, the total number of interlinkages is 14.4rr7A whence, 14.47T/A ~ 3B/X36 ~ 3B The other two paths are entirely outside of the conductors, 86 ELECTRICAL ENGINEERING and hence are acted on by all of the ampere-turns in the slot. These conductances are then, respectively, 0.47T ^ and 0.47r y:- Hi U The total effective conductance is then A . C . F per cm. length of armature, and the flux per slot is 6 = ING X 2.54? where I is the effective length of the armature, in inches. For both slots and 1 complete turn the inductance is N 2 X 6X3.2 X 11.7in.rA C F a ~ 10 8 7 10 8 Substituting numerical values for the slot and tooth dimensions, A = 1.233 B = 0.44 C = 0.213 D = 1.047 E = 0.51 F = 0.607, this inductance becomes L, = 449 [0.934 + 0.417 + 0.58] X 10~ 8 = 867 X 10~ 8 henrys. To this must be added the inductance of the end-connections. The flux produced by the end-connections per ampere-turn per inch of coil length may be taken as one-twentieth of that in the slot. It is, therefore, ^ X 3.2 X 1.931 = 0.309 lines. zo The coil divides as it passes out from the slot, so that only 3 conductors 3 conductors are grouped together. There- fore there are 37 amp.-turns producing flux around each conductor. If the length of the end-connections for 1 turn is assumed as 8 X diameter per pole, = 8 X 8 = 64 in., the flux surrounding each turn is = 0.309 X 37 X FIG. 63. 64 = 59.47 lines. The inductance is then T N 59.4 X 1 X 7 Le = TxW= 10* X 7 = 59.4 X 10- henrys. The total inductance is thus L = L, + L e = (867 + 59.4) X 10~ 8 = 926.4 X 10~ 8 henrys. ELECTRICAL CONSTANTS 87 The reactance of the short-circuited coil is X = 2irf c L = 6.28 X 540 X 926.4 X 10~ 8 = 0.0314 ohm, and the maximum e.m.f. of self-induction is E m = IX = 133 X 0.0314 = 4.17 volts. To overcome this e.m.f. the short-circuited coil is made to rotate in the field of the commutating pole. This field has been found to have an average density around the neutral axis of about 8620 lines per sq. in. In this case, the commutated coil has only one turn. Thus e.m.fs. are generated in one conductor under a " north" com- mutating pole, and in the other conductor under a " south" commutating pole. These e.m.fs. are similar, and together make up the total e.m.f. generated in the coil by rotation in the commutating field. The maximum value of this induced e.m.f. corresponds to the rate of cutting the flux in the center under the commutating pole. Consider a small distance, dx, Fig. 64, at this point. The flux through the area of width, dx, and average length, 11.7 in., of the iron in field and armature is FIG. 64. d = 8620 X 11.7 dx = 101,000te. The speed of conductors at the armature periphery is irD X r.p.s. = TT X 48 X 10 = 4807T in. per sec. The time required for a conductor to go the distance dx, is dT- dl 480ir , * induced = 10^ - J^_ = L53 V ltS J 4807T per conductor, or 3.06 volts per coil. This voltage opposes that due to self-induction, leaving as a resultant, 4.17 - 3.06 = 1.11 volts acting in the circuit. Since experience has taught that two volts potential difference can be taken care of by the resistance of the carbon brush no difficulties from sparking need be anticipated. CHAPTER XII DIRECT-CURRENT GENERATORS IN PARALLEL AND SERIES Shunt generators operate in parallel without the slightest diffi- culty. Generator No. 1 is first started and thrown on the line. Generator No. 2 is then brought up to about normal speed, the voltage is adjusted and the line switch is closed. Since genera- tor and line voltage are the same, no-load is taken by generator No. 2. By adjusting the field excitation of No. 2 the generator takes the desired share of the load. As its load increases its engine slows down, the governor opens and the speed is restored to normal. Series generators do not operate naturally in parallel. Assume, for example, that two series generators are in parallel, each taking its share of the total load. Suppose then that for some . . reason the voltage of No. 2 (Fig. 65) _ J j becomes slightly reduced. Its share of the load will fall off proportionately and, with this, its field excitation. Falling off of the field excitation further reduces the voltage and, con- sequently, the load, the excitation, and so on. The current is reduced to zero, then reversed in direction in both the field and the armature coils. The rotation of No. 2 remains the same, but the machine now acts as a series motor driving its engine. In practice, the rush of current dur- ing this period when the counter e.m.f. of generator No. 2 has been destroyed is so great that the circuit is opened by its fuses or circuit breakers. Series generators are not in common use, but this principle of instability in parallel operation applies equally to compound generators through their series field windings. With shunt generators there is no such instability. If the voltage of No. 2 falls off, its current likewise is reduced. But the effect of reduced current is to lessen the armature reaction, thus 88 DIRECT -C URRENT GENERA TORS 89 bringing up the voltage. The shunt field current is not affected since it is derived from the bus bars. Series generators and, more particularly, compound generators may be made stable in parallel operation by the use of an ' 'equalizer bus." This consists of a very heavy copper connection situated, as shown in Fig. 66, between the inner terminals of the series field circuits of the two (or more) generators. If, now, the voltage of No. 2 becomes reduced to a slight extent, current will flow from the + brush of No. 1 through the equalizer and into the series field coils of No. 2, main- taining the strength of the field of the latter. If the two generators, in normal operation, do F not divide the load prop- erly in the proportion of their respective ratings, this may be . corrected by inserting resistance in the series field circuit of that generator which takes too much of the load. The effect of the equalizer is to put the series field coils always in parallel. The voltage across these coils is therefore the drop between the positive brushes and the positive bus. The re- sistance of the equalizer is so low that its drop is negligible, so that the drop across all the series field coils is the same. Putting a shunt or diverter around one of the series field coils has no effect on the distribution of the load on any particular generator, as it affects all the series field currents alike, the proportions remaining the same. Direct-current Generators in Series. No inherent difficulty is encountered in connecting direct-current generators in series. Owing to the limited possibilities of constructing commutators that will permit the generation of very high voltages, where these are required in direct-current machines recourse is usually had to series connection. In electric railway work it is the general rule to employ both series and parallel connection of the motors to give flexibility in speed control. The Three -wire System. Two generators in series afford the simplest means of obtaining the three-wire system. This system, invented by EDISON, was devised to enable the use of large 90 ELECTRICAL ENGINEERING E numbers of low voltage incandescent lamps without, at the same time, entailing the use of a prohibitive amount of copper in the distribution system. As seen in Fig. 67, the voltage of the system is 2E t while that across any ele- ment of the system is only E. There are other ways by which power may be supplied to such a system. Thus, the source of power may be a single gen- erator of voltage, 2E, across whose termi- nals may be connected either a storage battery, as in Fig. 68, or two small gener- ators mounted on the same shaft, called a FIG. 67. balancer, and shown in Fig. 69. In either case the necessary condition is to have available some connec- tion point the potential of which is intermediate between those of the outer wires. The amount of current actually flowing in FIG. 68. FIG. 69. either the battery or the balancer set is small in case the two sides of the load are reasonably well balanced. Another scheme consists in the use of the three-wire generator. This is illustrated in Figs. 70 and 71. Fig. 70 shows a bi-polar machine constructed by reversing the windings on two adjacent FIG. 70. FIG. 71. poles of a four-pole generator. The potentials of the two brushes on the horizontal axis are the same and are midway between the potentials of the two other brushes. The object of making the machine bi-polar is to give an intermediate inactive DIRECT-CURRENT GENERATORS 91 belt along the commutator on which a brush may be placed without causing disruptive sparking. A better scheme is 'that of DOBROWOLSKY shown in Fig. 71. The armature is tapped at two opposite points which are connected, through slip rings, to a " choke" coil, which is simply an induction coil. This coil is wound upon a laminated iron core, and therefore is of high in- ductance. The e.m.f. impressed upon it is evidently alternating, and therefore very little alternating current can flow through the coil. The middle point of the coil must always be at a potential midway between those of the brushes. It may therefore be connected to the middle wire of the system. The disadvantage of using a battery is that some cells may be called on to supply more energy than others. It then becomes difficult to keep the battery uniformly charged, and deteriora- tion results. No such difficulty occurs with the use of balancers. They may be small, inexpensive machines, which when running idle take only a small current. As an example of the use of balancers and the economy of the three-wire system, consider the circuit illustrated in Fig. 72. 20 FIG. 72. The load consists of 40 amp. on the upper branch and 30 amp. on the lower. The system is therefore unbalanced. Currents and directions of flow are indicated for each portion of the circuit. The current in the middle or neutral wire varies, being 10 amp. in some sections and in others. Let it be assumed that the current required to run the balancer set is 1 amp. which would be indicated, if shown in the figure, by an arrow pointing down- ward in the balancer set. The current returning to the balancer over the middle wire is 10 amp. This current divides equally, 5 amp. flowing upward in balancer A, combining with its down- ward flowing 1 amp. to give 51=4 amp. in A, and 5 amp. 92 ELECTRICAL ENGINEERING flowing downward in B, combining with its downward flowing 1 amp. to give 5+1=6 amp. in B. Current in A flows similarly to that in the main generator. Thus A acts as a generator, supplying 4 amp. to the load. Current in B flows in the opposite direction; thus B acts as a motor and drives A. The difference in current between that in B and that in A is 2 amp., which, when multiplied by E, the voltage across B } gives 2E, the power required to drive the balancer set. If the generator voltage be assumed as 200 (that is, 2^ = 200), then the generator output, or rating, if this be full-load, is 200 X 36 = 7.2 kw. The balancer, A, rating, as a generator, is 100 X 4 = 0.4 kw. ; the balancer B, as a motor, receives input = 100 X 6 = 0.6 kw. The line drop from the generator to the load is (40 -+- 30) r = 70r, where r is the resistance of each of the outer wires. The line loss, in transmission, is (40 2 -f-30 2 )r = 2500r. If the entire load were on the two-wire system, the current in each wire would be 70 amp., the line drop, using the same size wires, would be 140r, and the line loss would be (2X 70 2 )r = 9800r. Comparing the two-wire system, using the same size of outer wire, drop, three-wire _ 70 _ drop, two-wire ~"~ 140 loss, three-wire _ 2500 _ loss, two- wire ~ 9800 The middle wire, carrying 10 amp., has no effect on the total drop between the outer wires. It does have some effect in slightly unbalancing the voltage of the two branches of the system. Thus, assuming the voltage across the two machines of the balancer to be exactly equal, which is very nearly true, and taking this voltage as E, the voltage across each branch of the load may be found. Across the upper branch it is, E - 40r - lOr = E - 50r. Across the lower branch the voltage is E - 30r + lOr = E - 20r. The amount of unbalancing of the voltage is therefore (E - 50r) - (E - 20r) = 30r. DIRECT-CURRENT GENERATORS 93 To get a concrete idea of the amount of this unbalancing, let SO the line drop, 70r, = 10 per cent. Then 30r = ^ X 0.1 = 0.043 = 4.3 per cent. When the load consists of lamps it is necessary that the two branches shall be sufficiently well balanced to prevent excessive variation in voltage. This is usually very easily accomplished. The middle wire adds, directly, a small amount to the line loss. In this instance, the loss in this wire is 10 2 r = lOOr. The total loss in the system is therefore 2600r, and the ratio 2600 of losses of the two systems is QQnn = 0.265. Where the percentage line drop or the percentage line loss is specified, and must be the same with either system, the ad- vantage of the three- wire system is in the saving in the cost of copper. On that basis, let the calculations as already carried out for the three-wire system be assumed as fulfilling the re- quirements, that is, Line drop = 70r. Line loss (two-wire) = 2500r. The two-wire system, to give equal line drop must be com- posed of wires determined by the equation, 2 X 70 X r' = 70r, where r' = resistance of one wire of the two-wire system, and r, as before, is the resistance of one of the outer wires of the three- wire system. Then and each wire of the two-wire system will be twice as large as each outer wire of the three- wire system. Assuming the middle wire equal to the outer wire, the two-wire system will require four-thirds as much copper as the three- wire system. Since, however, the variation in voltage . is felt by all the lamps on the two-wire system, while on the other system ap- proximately one-half the variation in voltage is felt by each branch, it is more reasonable to calculate on the basis of equal percentage drop in the two systems. 94 ELECTRICAL ENGINEERING Percentage drop, three-wire, = = 35 - For equal percentage drop, therefore, the two-wire system will require eight-thirds as much copper as the three-wire system. On the basis of equal power loss in the outer wires, 9800r' = 2500r, ' r ~ 9800 " Adding the middle wire, equal to an outer wire, the two-wire system will require n 055 _u n 12Y5 = 2.61 times as much copper as the three- wire system. Problem 31. What saving in copper does the three- wire system give over the two-wire system, when the load is balanced, on the basis of (a) equal percentage line drop, (&) equal line loss? 1. Middle wire equal to outer wire. 2. Middle wire one-half of outer wire. 3. Show that with a balanced load no current flows in the middle wire. Problem 32. One hundred 60-watt tungsten lamps are to be supplied with power at 3 per cent, line loss. The line length is 600 ft. Lamp voltage is 120. The neutral wire is to be one-half the cross-section of each outer wire. Find the size of the required wires, and show that the weight of copper is approximately 190 Ib. Show that on the two-wire system 610 Ib. would be required. Feeder / T f Trolley Wire 4 rf / E5 iiUj Rail EEL COJQ FIG. 73. Boosters. Generators are frequently connected in series for the purpose of regulating the voltage and equalizing it along a line in which there is considerable voltage drop. Fig. 73 shows a simple arrangement of a street railway circuit in which a booster is used. The generator, G, supplies power DIRECT-CURRENT GENERATORS 95 to the system, including that delivered directly to the car and that used in driving the motor M . The motor and booster form, usually, a directly connected set. One terminal of the booster is connected to the trolley wire at the station, the other is con- nected through a heavy feeder to some distant point on the trolley wire. As an example of the effect of using a booster, consider the following : Problem 33. A trolley line 3 miles long is supplied with power by a generator at 600 volts. The trolley wire is of No. 00 B. & S. wire, having a resistance of 0.4 ohm per mile. The rail return has a resistance of 0.05 ohm per mile. A feeder, consisting of three No. 0000 B. & S. wires, of 0.087 ohm per mile, extends from the station to a point 2 miles distant, where it connects with the trolley wire. The booster voltage is maintained at 40. Find the voltage on the car as it proceeds from the distant end of the line toward the station, assuming that the current taken is at all times 200 amp. Solution. It will be of interest, first, to determine the voltage on the car at the distant end when the booster is disconnected. Drop in the trolley wire is 200 X 0.4 X 3 = 240 volts. Drop in rail = 200 X 0.05 X 3 = 30 v. volts. .'. Voltage on car without booster = 600 - 270 = 330. This is to illustrate the necessity of doing something to improve the regu- lation of the line. With the booster connected, the problem becomes one for the application of KIRCHOFF'S laws. The circuit is represented diagrammatically, in Fig. 74, for the case of the car at the end of the line. Arrows indicate arbitrary directions of flow of current. Let the voltage on the car be denoted by E. By KIRCHOFF'S laws, ii + 12 = 200 and 40 - ; 2 r 2 + iiri = 0, whence, eliminating i\ between the equations, 200r 1 + 40 iz = : 74 Substituting values, ri = 0.4 X 2 = 0.8 ohm r 2 = 0.087 X 2 = 0.174 r 3 r = 0.4 = 0.05 X 3 = 0.15 200 X 0.8 + 40 -^8Tol74-= 2 200 - 205 = - 5 amp. 96 ELECTRICAL ENGINEERING The equation of the mesh composed of the generator, ri, r 3 , E and r is 600 = iiri + 200r 3 + E + 200r . Substituting values and solving for E, E = 600 + 4 - 80 - 30 = 494 volts. Thus, there is a total drop of 106 volts instead of 270 volts without the booster. Now let the car be at the point, O, where the feeder joins the trolley wire. Evidently the same equations hold, and z' 2 = 205 amp., ?i = 5 amp. r is now 0.05 X 2 =0.1 ohm. The mesh equation is now 600 = iiri + E + 200r . /. E = 600 + 4 - 20 = 584 volts. When the car is at a point 1 mile from the generator, the current and voltage equations are : ^ + 12= 200 and 40 - i, (r, + + ti = 0. Solving for i 2 gives i = 123 amp. and ii = 200 - 123 = 77 amp. The mesh equation of voltage is 600 = ii ^ + E + 200r , where r is now 0.05 ohm. /. E = 600 - 31 - 10 = 559 volts. Thus, it is seen that, by this simple connection of the booster to a point chosen more or less at random, the voltage has been rendered much more nearly uniform than it would be without the booster. Problem 34. As a further study of the booster problem, consider that in the above case the feeder is to be connected to the trolley wire at two points, namely, at 1.5 and 2.5 miles from the generator. Find the voltage on the car at each half-mile point, and plot against distance from the generator. CHAPTER XIII DIRECT-CURRENT MOTORS If two shunt generators connected in parallel supply power to a certain load, as in Fig. 75, the division of the load between the generators will depend upon their respective degrees of excitation. By weakening the field of No. 1, it will take less of the load until, by continued weakening, it takes none at all and finally receives current from No. 2, thus being run as a motor. With a change in direction of flow of current in the armature comes a change in the direction in which the armature tends to rotate due to its current, the direc- tion of the field remaining constant in shunt machines. As a generator the rotational force of the armature is counter to the actual direction of rotation which is , ., , . . . TT FIG. 75. due to tne driving engine. However, the actual direction of rotation does not change when the machine ceases to act as a generator and becomes a motor. With the series generator, reversal of the armature current also reverses the field. To obtain a generator action from a series motor, therefore, requires reversal of rotation. It has been shown that, with generators, a forward shift of the brushes increases the armature demagnetization. With a shunt motor the armature currents are reversed, the armature ampere-turns are reversed, and the effect of the arma- ture, in shifting the resultant flux, is consequently reversed. Therefore, the brushes of a motor require to be given a backward shift. The effect of a backward shift on a motor, like the for- ward shift on a generator, is to increase the armature demag- netizing ampere-turns. With direct-current motors, the impressed e.m.f. is the sum of the counter e.m.f. and the ir drop. Thus, the fundamental equation is E = Ei + ir 7 97 98 ELECTRICAL ENGINEERING where E = impressed e.m.f., Ei = counter e.m.f., i = current, and r = resistance of armature, brushes, etc. The generator equation (20) also applies to the counter e.m.f., since the counter e.m.f. is the generated e.m.f. of the motor due to the rotation of its armature conductors in the field. where , 4< = 105' Substituting this value of E i} E = kf + ir (24) whence f = frequency. To transform frequency to speed, r.p.m. p J ' 60 *\i* where p = number of poles. For ordinary operation, pi f = 7' approximately. There are three ways of changing the speed of a direct-current motor: (1) by changing E, the impressed voltage; (2) by changing by means of a field rheostat ; (3) by changing by shifting the brushes. Shifting the brushes is not an effective means of speed regula- tion since it introduces trouble from sparking at the brushes. Types of Direct-current Motors. The principal types of direct-current motors are known as shunt, series, cumulative- compound, in which the series and shunt turns act in the same direction, and differential-compound, in which the two field m.m.fs. are arranged to oppose each other. DIRECT-CURRENT MOTORS 99 Speed Characteristics of Direct-current Motors. These are curves between speed and load, the latter being the independent variable. To determine the effect of load upon speed, in the case of shunt motors, it is seen from Eq. (25), _ E - ir * = ~ that an increased ir drop tends to reduce the speed. It has also been shown that is reduced by armature reaction, in pro- portion, roughly, to the load. Therefore, for shunt motors, the relation between the armature reaction and the ir drop will de- termine whether the motor will speed up or slow down with an increase of load. In general, if the magnetization of the field extends above the knee of the saturation curve, the motor will slow down, while below the knee the motor will speed up. Evi- dently, a degree of magnetization might be obtained which would result in practically constant speed. The cumulative-compound motor slows down with increase of load, since the effect of the series turns is to strengthen the field. The differential motor speeds up with increasing load, due to the op- position of the series and shunt field m.m.fs. The series motor speed is governed almost entirely by its field, which is nearly proportional to the load cur- rent. At light loads, the speed be- comes high and the operation of the motor is unstable. In Fig. 76 is shown a set of speed characteristic curves. The student should be able to establish the general speed equations and derive curves for each type of motor. Power and Torque. Power input to the motor is obtained by multiplying Eq. (24) by i, thus Wi = Ei = Ej + i*r = kf<(>i + i 2 r. In this, equation E# represents the output of the motor in mechanical work, including bearing friction and windage; i 2 r is the power lost as heat developed in the armature. Expressed in horsepower, the output is Load FIG. 76. 100 ELECTRICAL ENGINEERING Horsepower may also be expressed as 2irRnF P' ~ 33,000 where R = radius of armature in feet, n = revolutions per minute, F = force in pounds on the armature conductors. 2wn = co = angular velocity, and RF = T = torque. Thus and 746 33,000 33,000 Ed 27rnX746' But output is also, by (24), kfi. . ^ 33,000 kfoi = 2irn X 746 ' Also, since / = T = where p = number of poles, 0.0587 kp4>i = 120 33,000 kpi 27rX 746X120 This expression may be reduced still further, since Thus, for where t = number of turns in series on the armature. a motor of p poles and t turns in series, T = 0.2348 tpi X 10~ 8 ft.-lb. Torque Characteristics. From the above equation of torque it is possible to construct curves showing torque variation with load current. It is necessary, however, to be able to find the value of in each case. With shunt motors is nearly constant, and torque is therefore nearly proportional to current. With series motors increases with i, and torque therefore goes up as the square of the current, approximately. l Fig. 77 gives a set of torque characteristics for the four types of direct-current motor. 1 When the field core becomes saturated, increase of current does not produce much increase of flux. Under heavy loads, therefore, the torque of a series motor increases more nearly in direct proportion to the current. Current FIG. 77. DIRECT-CURRENT MOTORS 101 Problem 35. Direct-current motors and generators being entirely similar as respects fundamental equations, armature reaction, etc., it is thought best to submit to the student the problem of the direct-current shunt motor instead of presenting it here in detail. Let the generator whose design was worked out in Chap. X be now considered as a shunt motor. The series turns will then be disconnected. With 250 volts im- pressed on the armature and maintaining constant shunt field amp. -turns of 7500, let it be required to calculate the speed and plot its values against those of the load current. Choose current values of 0, 1000, 2000, 3000 amp. Assume a constant brush shift of 15. The fundamental speed characteristic, Eq. (20), has been found to be E - ir '--*r- number of poles X r.p.m. pn where 2X60 = 120' E = impressed voltage, r includes both armature and brush resistance. where t = number of turns per pole on the armature and $ is the flux cutting the armature conductors. For this last it is sufficiently exact to assume $ at load amp. -turns at load at no-load ~~ amp. -turns at no-load (See armature reaction, Chap. X.) Problem 36. The same problem as the preceding should now be worked out, using (1) E = 270 volts, (2) E = 220 volts. Question. What, in general, is the effect on shunt motors of increasing or lowering the terminal voltage, as regards (a) speed, (6) torque, (c) output, (d) efficiency? Problem 37. Let the above motor be calculated as a differential-com- pound machine, the series ampere-turns to be so adjusted as to give the same field strength at full-load as at no-load. Plot speed vs. armature current for impressed voltage E = 250. Problem 38. Same as 37 only the motor is to be connected as cumulative compound. Problem 39. If, now, the entire field strength were determined by the series turns, so that at full-load there should be 10,427 series amp.-turns, 1 calculate and plot the speed for variation of load. Series field circuit resistance may be taken as 0.00134 ohm. Problem 40. In problems 35, 37, 38 let the speed be maintained con- stant by variation of the shunt field current. Let this speed be that of the shunt motor at no-load (E = 250). Plot curves between field current and load current. Problem 41. Show how to obtain constant speed by shifting the brushes, and work out numerically, as far as possible, the case of the shunt motor. Plot a curve between degrees of brush shift and load current. 1 Same as required for the generator at full-load, Chap. X. CHAPTER XIV N FIG. 78. THEORY OF THE BALLISTIC GALVANOMETER This particular type of galvanometer is of importance in magnetic measurements, especially in the determination of the hysteresis loop. It consists, usually, of a coil of fine wire wound upon a steel cylin- * der, freely suspended between the poles of a magnet as illustrated in Fig. 78. It has been shown that the force exerted on a wire carrying current, when placed in a field perpendicular to the lines of flux, is F = Eli dynes, where i is current in abamperes, I is length of wire in centimeters and B is flux density in lines per square centimeter. If the wire is one side of a rectangular loop, then the turning couple of the loop is C = 2pBli dyne-cm. When the loop is displaced by an angle, 0, from the direction of the flux lines (Fig. 79), the couple is C = 2pBli cos 6 = ABi cos 6, where A = 2pl = area of the loop. When the current is sent through the loop, the action of the couple produced is to turn the loop through an angle, 0. In order to oppose this action, a spring is so attached to the loop as to introduce an op- posing couple, k&, which balances the swing of the loop. Then ke = ABi cos and K0 AB cos 6 where k is a constant of the spring. 102 Force FIG. 79. THEORY OF THE BALLISTIC GALVANOMETER 103 For small angles, 6 = sin 6. Substituting this, k sin 6 k cos tan Thus, the current in the loop is directly proportional to the tangent of the angle of deflection; hence, the " tangent" galvanometer. For small angles, also, 6 = tan 6. Thus, the galvanometer may be used as an ammeter to measure directly the current, so long as the angle of deflection is kept small. In the ballistic galvanometer the moving part is designed to have much inertia, so that its natural period of vibration shall be long in comparison with the time of change of the flux to be measured. Thus, a change of flux, produced in a sample of iron under test by altering the number of ampere-turns on the iron, will take place before the loop can move, that is, while = and cos 0=1. The couple on the loop is then C = ABi, which causes the loop to accelerate. Therefore, ABi is the couple of angular acceleration, and where /o = moment of inertia of the moving element, and o> = angular velocity. But idt is the quantity of electricity flowing in any time, dt. Therefore the total quantity ... /o | , h <*> J ldt = AB J d " = ~KB (26) AD where a> is the final velocity attained. The deflection is, however, limited by k0, the torsion of the spring. The work done in overcoming this torsion is then W = kOdd = where is the maximum deflection. = (k 104 ELECTRICAL ENGINEERING Solving this equation, I/O t/o ^OA/ But by (26), 0)fl Substituting this value of co , whence, Q = ^^ (27) In this equation all terms are constant except , the maximum deflection of the loop. If, now, the change of flux is d in a time dt, the e.m.f . induced in the coil surrounding this flux is N_ d$ _ . 10 8 dt '' IT) where N is the number of turns of the coil, r is the resistance of the circuit, and i is the current set up in the circuit. Then, transposing, Nd idt = - r X 10 8 ' The total quantity of electricity set flowing by the change of flux is then r X 10 8 L r X 10 8 ' whence, from (27), Qr X 10 8 _ 6 r X 10 8 \/J7fc N ABN is the maximum value of the flux. There is thus a direct relation between flux and maximum deflection, and 0o is therefore a measure of the flux. CHAPTER XV VECTOR REPRESENTATION OF ALTERNATING- CURRENT WAVES In Chap. VIII the graphical relationships of the waves of voltage and current in an alternating-current inductive circuit have been developed, and the values and meaning of average and effective values of a sine wave have been discussed. The waves of Fig. 37 may also be represented as vector pro- jections of their maximum values on the vertical axis, as shown in Fig. 80. Since i = I m sin the length of ^ r the current vector is taken as I m and the value, i, at any instant, is the vertical pro- jection of I m as it uniformly rotates, at speed 2-7T/ about the origin. The vectors all have FIQ ^ the same speed of rotation so that their re- lations to each other are constant. Hence their position in space at any desired instant may be chosen. Let that instant be when = o, in Fig. 37. Then i = I m sin = o, and I m must be laid off horizontally. rl m , the maximum value of the e.m.f. consumed by the resistance, since it is in time-phase with I m , is also laid off horizontally; xl m , the maximum value of the e.m.f. consumed by the inductive reactance, x, is 90 ahead of I m , and is therefore laid off vertically upward. Thus xl m is positive maximum when I m is at zero, becoming positive. rl m and xl m may now be added vectorially, giving ZI m or E m which is the maximum value of e. E m is seen to be placed at an angle a ahead of I m , such that tangent a = -y = J- m ' This relation is also of fundamental importance. The numerical value of E m is obtained by the relation E m = Vl m 2 r* + I m V = I m Vr 2 + z 2 The quantity \/r* + x 2 is called the impedance and is denoted by the letter z. 105 106 ELECTRICAL ENGINEERING Problem 42. Draw the vectors of e.m.f. and current of problem 28, Chap. VIII, and show that the angle of lag of current behind e.m.f. is 38 40'. In the representation of waves by vectors, the vectors are not, in reality, moved, but their relative positions in space are con- sidered. Since no rotation is required, they may therefore be drawn in length equal to their effective values, and this is the common method of representation. Also, since (7 m z) 2 = (I m r)* + U m x) 2 (28) 2 = r 2 _|_ 3.8 (29) and the vector relationship holds for (29) as for (28). There can be constructed what is known as the im- pedance triangle, Fig. 81, in which a = r, b = b x x, c = z, and tan a = - = Thus, FIG ' 8L x = z sin a. Substituting these values in (19), e = I m (r sin 6 + x cos 6), gives, e = 7 m 2(sin B cos a + cos 6 sin a) = I m z sin (8 + a), or, substituting for 6, its equivalent, coZ, e = I m z sin (ut + a), (30) in which at is a variable angle depending on t, and a is a constant angle determined by the relative values of x and r. Eq. (30) shows that e, like i, is a sine wave quantity, but that there is a constant angular or phase difference, a, between them, a is called the angle of lead or lag, depending on whether it is posi- tive or negative. The relations indicated in Fig. 81 may also be expressed by the notation of complex quantities. Thus, c = a -f jb. The addition of the letter j to the equation simply means that the quantity, 6, is to be drawn vertically upward. If it were j, b would be drawn vertically downward. A dot is put under ALTERNATING-CURRENT WAVES 107 c which means that c is dealt with as a vector quantity. Without the dot, the scalar or numerical value of c, only, is meant. Thus, _ c = \/a 2 + V. Problem 43. Show graphically that 3-J3 is a vector of length 4.24, which makes an angle of 45 with the horizontal axis. Show that a vector of length 12, at angle 120, is represented by the expression, 6 + j 10.4. Calculate and draw the following vectors : c = 3 j'2, c = 4 + j, c = -2 + j3, c = -4 -J2. j also means a rotation of 90 in the positive or counterclock- wise direction. If the vector, a, is multiplied successively by j, several times, its direction is shown as follows: Vector Angle .... 90 J u jja a, 180 270 iiiia = a. . . . 360 = Thus may be written, whence or, j is identical with i, used commonly in mathematics to denote imaginary quantities. 1 If it is desired to rotate a through 30, we can write a = a cos 30 -f j a sin 30. To rotate correspondingly, ^~~ ' a 8in30 a = a cos a + j a sin a, / o i o\ FIG. 82. = a (cos a -f j sin or). Suppose a is first rotated 30, then 60 more. Then a = a (cos 30 + j sin 30) (cos 60 -f- j sin 60). Problem 44. Prove that this double rotation results in a = ja Consider the simple case of alternating current in an inductive resistance, Fig. 83, where current, /, resistance, r, and reactance, 1 In electrical engineering j is used instead of i, because i is used to denote current. 108 ELECTRICAL ENGINEERING x, are known. / is chosen as the zero vector. Then I = i. Frequently it is well to choose as the zero vector, or vector drawn at 0, some known quantity. In order to determine the positions of the vectors of electromotive force, etc., with respect to the zero vector, there are two rules, previously brought out, which are important to remember: Rule I. The e.m.f. consumed by resistance is in time-phase with the current, and in the same direction. Rule II. The e.m.f. consumed by inductive reactance is in time-phase 90 ahead of the current By these rules may be drawn the vector diagram, Fig. 84, in which the vector sum of ix and ir is iz, which is the total electro- motive force consumed. FIG. 83. FIG. 84. This electromotive force consumed, or vector E, numerically equal to iz, is represented by the relation E = ir + jix i(r + jz). The impedance is thus expressed as r + jx, and it is a vector of magnitude z = \/r 2 + x 2 , and the angle between the impedance and the resistance is defined by the relations. and Z COS a tan a = - r The e.m.f. consumed in the circuit is, in general, E = IZ = (iji'} (r+jx). The current may or may not be chosen as the zero vector. If it is so chosen, / = i. If not, then / = i ji' } where i' is the wattless component of the current. The impedance is always z = r + jx. Assigning positive or negative values to the wattless component i f , we may write, in any case, I = i+ ji'. ALTERNATING-CURRENT WAVES 109 It should be remembered that a leading component requires a + sign, and a lagging component requires a sign. Therefore, E = IZ = (i + ji') (r + jx) = ir i'x + j(i'r + ix) If the current is taken as the zero vector, then E = i(r + jx) In the general expression (31), an arbitrary zero line is chosen, as in Fig. 85. In the simpler case (32), the direction of / is chosen as the zero line, whence I i and i' = 0, and the vector diagram becomes that of Fig. 86. (31) (32) ix FIG. 85. %r FIG. 86. Problem 46. One ampere flows in a circuit of 1 ohm resistance and a vari- able reactance. Plot curves of Ir, Ix, Iz drops and phase angle against x, when x varies from to 5 ohms. Take / as the zero vector. Then 7 = i = 1. Solution. Tabulating : ' tan a X 0.5 1 2 3 4 5 ix 0.5 1 z 1 1.12 iz 1 1 12 x 5 r - a 26 35' The blank spaces may be filled in by the student. Consider the same case, Fig. 79, but with E known and / unknown. E, then, may conveniently be chosen as the zero vector, and I = e _ = e e(r-jx) z r + jx ( (33) 110 ELECTRICAL ENGINEERING The last expression of (33) is obtained in accordance with a third rule, as follows: Rule HI. Never allow an equation to remain with a complex denominator. Thus (33) becomes where g (34) (35) FIG. 87. g + jb F, is called the admittance-; g is called the conductance, and b the susceptance of the circuit. The diagram of currents may now be drawn to correspond with Fig. 87, for e.m.fs., in which eg is the com- ponent of the current in phase with e, that is, it represents energy expended, and eb is the component 90 behind e, called the reactive or wattless component because it does not represent any expenditure of energy. The power input to the circuit is then Power input = e X eg = e z g, and this is found equal to I 2 r. The numerical value of the current = I = e\/g 2 + 6 2 . Problem 46. Let E = e = 1; x = 1; r varies from to 10. Plot curves of / vs. r, and I 2 r vs. r. Calculate the maximum value of the power loss and find the value of the resistance which gives the greatest dissipation of power. Plot the 3 current waves, that is, the power current, eg., wattless cur- rent, eb, and total current, ey, for the condition of maximum power loss. Solution. e 1 1 Vr 2 -f x 2 \/l + r 2 * - r x 1 v = - z ', = --,; y = - Tabulating : r 2 4 6 8 10 Z 1 2.27 4.12 7 1 0.44 0.242 / 1 194 7r.. '..... 0.388 * ALTERNATING-CURRENT WAVES 111 Power, W = Pr = - z r *?-.. For maximum power -p- = 0. .'. x 2 r 2 = 0, and r 2 = x 2 . .'. r 2 = 1, r = 1, and W = 1 X j-qjj = 0.5 watt. To get current waves for maximum power loss, then r = 1; x = 1; Z = 1.41. e _ _e(r - j x) _er , ex _ where ^ = -^- 2 and 6 = ^* The effective values of current are, therefore, eg = 1 X K = 0.5, in phase with e, jeb =-je~=-jlX^ = -jO.5, or 0.5, 90 behind e, e(g +jb)=eY=^ = ^ - 0.707, lagging behind e, by an angle tan" 1 & V2 Maximum values are ^ TO = \/2^ = 1.41, (Eg) m = 0.707; (J^6) w = 0.707; (EY) m = 1. Circuit of Resistance in Series with an In- ductive Impedance. The impressed e.m.f., E, of the circuit is known, also the resistance, E r, and impedance, Zi = r l + jxi (Fig. 88). # I is taken as the zero vector. Then, FlG 88 where r , Xi Q v~2> o = -- ^- 2 ; Z and r = r -j- ri. The drop across the impedance, Zi, is #1 = 7 Zi = e(g + j6) (7*1 + ji (s^i + n. where a = 0n 60; ij 6 = gxi + 112 ELECTRICAL ENGINEERING Problem 47. In the above circuit, Fig. 88, let E = 10, r = 1, n = 0.5, Xl = 2. Draw the vector diagram and waves of e, EI and /. Circuit of Two Inductive Impedances in Parallel. Let E, Z\ and Z 2 be known (Fig. 89). To determine 7, /i and /2. FIG. 89. We have: /i = eY l} J 2 = I -li + J, -a(Fi-+ 7,). ' Or, / 2 = e(flfi +j'6i)i 62) = e(G where G = 0i + g t , B = 61 + 62. D Tan a -^-> gives the phase relation of / and e. Problem 48. In the circuit of Fig. 89, let E = 1, n = 1, x l = 0.5, r 2 = 2, x 2 = 2. Draw vector diagrams and waves of E, Eg, Eb, and /. CHAPTER XVI THE SYMBOLIC METHOD IN TRANSMISSION LINE CALCULATION KENNELLY AND STEINMETZ have introduced the so-called symbolic method of representing electrical relations. This method is neither vector analysis nor quaternions, but is in many ways similar to both. It enables the use of simple algebraic transformation when dealing with vector quantities of the same rate of rotation or frequency. Thus, it is directly applicable when, for instance, multiplying a current by an impedance, since the resultant e.m.f. is of the same frequency as the current. But when multiplying current and e.m.f., it is applicable only after some modification, since the product represents power, which is a vector of double frequency. Addition. Let there be two vectors, a\ + J&2, and bi -\- jb 2 , and let their sum be a vector C. Then, C = ai + ja 2 + 61 -f jb 2 = ai + bi + j(a 2 + 6 2 ) = ci + jc z , where Ci = ai + bi and c 2 = a 2 + b z . Multiplication. We have, evidently, Oi + ja 2 )(6i + J6 2 ) = ai&i - a 2 b 2 + j(a^ -f bia z ) = di + jfa, where di = dibi a^bz', dz = Q>ib% + b\a,z. In general, if i + ja z = bi + jb 2 then ai = 61 and a 2 = 62- Power. At any instant, p = ei, where e and i are instantaneous values of voltage and current. In the case of sine waves, where e = E m sin ui and i = I m sin (* + ), p = ei = E m l m sin cot sin (co + ) 8 113 114 ELECTRICAL ENGINEERING Problem 49. Plot waves of voltage and current, and by multiplying their values at certain instants along the curves show that the resulting power curve is a sine wave of double frequency. Let E m = 1.4, Im = 0.7 (1) a = (2) a = 45 (3) a = 90. Fig. 90 shows the curves plotted for the case of a. = 0. The energy developed in the circuit, in any time dt is pdt. The total energy during a cycle is then J] T pdtj where T is the time of one complete cycle. But this is the - area enclosed by the power curve and axis, shown shaded. As the values of power are always posi- FIG. 90. tive, the area represents energy expended, or work done. The student should show that when a is not 0, there is also negative power, which represents energy returned to the source, the total energy expended during a cycle being the difference between the positive and negative areas enclosed by the power curve and the axis. Average Value of Power during a Period. -r-This will be, p m l m sin ut sin (coZ -f a)dt, which, the student should show, is cos a. This may be written E m I, V2 V2 COS a = El cos a (36) where E and I are effective values as usual. Thus the important result is found, that, in case of sinusoidal current and voltage waves, the average power is equal to the effective value of the current times the effective value of the voltage into the cosine of the phase angle between the two. This is illustrated in Fig. 91, and it is seen that when 7 is zero vector = i, P = El cos a = el = ei. Similarly, when E is zero vector = e, P = Ei = ei. FIG. 91. METHOD IN TRANSMISSION LINE CALC ULA TION 1 15 Power is obtained by multiplying either quantity by the pro- jection of the other upon it. In general, if E makes an angle 7, and 7 an angle with the zero axis, where a = 7, Therefore, 0> 7 . P = EI cos a = #7 cos (0 - 7) = EI (cos cos 7 + sin sin 7) (36) t e But cos cos 7 == sin/3 = sin 7 = Substituting these values in P = ei + e'i' which is the general expression for the average power. Example. Let E = e + je' i = i -f ji r . Then by (37) P = ei + e'i'. Suppose, however, that we carry out the multiplication of the vectors. Thus, EI = ( e +je')(i+ji f ) = ei e'i' + j(ei' + e'i) The numerical value of this expression is (36) (37) FIG. 92. \(t- eY) 2 + (e'i- ei') 2 which is obviously not the same as (37), neither is its real com- ponent the power, since it has a minus sign. It has been shown in Fig. 90, that power is a quantity of double frequency. It can therefore have no phase relationship with E or 7. Hence, in the case of power or any double frequency quantity, the operation of multiplying single frequency quanti- ties is inadmissible. On the other hand, it is known that the product (i-ji>) (r+jx) = E is quite correct, since the fundamental frequency only is involved. 116 ELECTRICAL ENGINEERING The operation of obtaining the power from two vectors E and I, is called " telescoping" the vectors. Thus, the prod- uct of the "real" components is added to the product of the "imaginary" components, without any change of sign due to the presence of j. Power Factor. In the expression for power (36) the term cos a is called the power factor. The product El represents true power only in certain special cases, particularly with direct currents. T r v j ^ j .- true power Power factor may be defined as the ratio - ' apparent power where apparent power = El. El is also called the volt-amperes. Since El cos a is the true power, . El cos a power factor = p.f. = ^7 = cos a. Jiil Transmission Line Calculation. The calculation of circuits may now be continued to include the case which represents a simple transmission line possessing concentrated resistance and inductive reactance, being supplied with power at one end by a generator, or source of alternating current, and terminating at the other end in any prescribed load. A cus- tomer usually desires constant volt- FlG 93 age, E, at the load. In Fig. 93 is shown a generator supplying power over a transmission line of impedance, Z = r + jx t at voltage E Q to a load, the current of which is i + ji' at voltage E. E and i are in time phase ; E and i f are in time quadrature. (1) Let E be known, and be taken as the zero vector, = e. Then, the voltage at the generator terminals, Eo = e + IZ = e + (i + ji') (r + jx) = e -f ir + ijx + ji'r i'x = e + ir i'x + j(ix + i'r) = a -f- jb, where a = e -f ir i'x, b = ix + i'r. METHOD IN TRANSMISSION LINE CALCULATION 117 The power factor of the load, cos a = -=. = y Generator volt-amp. = IE Q . Power of Generator. P* Q = E I cos 7, where 7 is the angle between E Q and 7. Vector relationships are shown in Fig. 94: (a), for leading and (6), for lagging current. In the first case 7 = a |3 is the angle between E Q and 7. /. Cos 7 = cos (a |8) = cos a cos + sin a sin /3. i i' But cos a = j, and sin = y-' a 6 Likewise, cos /5 = yr and sin = ^- /. Cos 7 = yyr (m + i'6). Substituting this value into the equation for power of generator, PQ = E I COS 7 = itt + *'6. P could be more quickly obtained by simply telescoping the vectors a + jb and i -f ji'. Power factor at the generator power PQ Ei Efficiency of transmission = ^5 -T output Apparent efficiency == ratio, Regulation = -- ^ -- Having obtained the general expression for the various quan- tities which enter in, we may now take a specific example of transmission line calculation. Problem 60. In Fig. 93, let E = 1, r = 0.1, x = 0.2, i = 1. Let i' vary from 1 to +1. Determine all the quantities, i.e., current, generator voltage, volt- amperes, power factor, power, transmission efficiency, apparent efficiency and regulation. 118 ELECTRICAL ENGINEERING Tabulating: 1 j' '. -1.0 -0.75 -0.5 -0.25 0.0 0.25 0.5 0.75 1.0 If 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 i'x 0.2 0.15 0.1 0.05 * 0.0 -0.05 -0.1 -0.15 -0.2 a 1.3 1.25 1.2 1.15 1.1 1.05 1.0 0.95 0.9 ix 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 i'r -0.1 0.075 0.05 0.025 0.0 0.025 0.05 0.075 0.1 I 0.1 0. 125 0. 15 0. 175 0.2 0.225 0.25 0.275 0.3 a j . 1.69 1.56 1 .44 1.32 1 .21 1.10 1.0 0.90 0.81 6 2 0.01 0.0156 . 0225 . 0306 0.04 . 0506 0.0625 0.0756 0.09 a 2 + 6 2 1.7 1 . 5756 1.4625 1.3506 1.25 1.1506 1.0625 0.9756 0.9 -y/ 2 4. b 2 .. . . 1.31 1.25 1.21 1.16 .12 1.08 1.03 0.985 0.95 Eo 1.31 1.25* 1.21 1. 16 .12 1.08 1.03 0.985 0.95 i' 2 1.0 0.56 0.25 0.0625 .0 . 0625 0.25 0.56 1.0 i 2 + i' 2 2.0 1.56 1.25 1 .0625 .0 1.0625 1.25 1.56 2.0 V* 2 + i' 2 ..-. 1.41 1.25 1.12 1.03 .0 1.03 1.12 1.25 1.41 I 1.41 1.25 1.12 1.03 .0 1.03 1.12 1.25 1.41 E I 1.85 1.56 1.36 1.23 .12 1.11 1.15 1.23 1.34 i/I 0.707 0.8 0.895 0.94 .0 0.94 0.895 0.8 0.707 CM 1.3 1.25 1.2 1.15 .1 1.05 1.0 0.95 0.9 bi'. 0.1 0.0937 0.075 0.0437 0.0 0.0563 0.125 0.206 0.3 Po 1.2 1.156 1.125 1.106 1.1 1.106 1.125 1.156 1.2 i'/i -1.0 -0.75 -0.5 -0.25 0.0 0.25 0.5 0.75 1.0 tan a -1.0 -0.75 -0.5 -0.25 0.0 0.25 0.5 0.75 1.0 a 45 37 26 30' 14 14 26 30' 37 45 Cos a 0.707 0.8 0.895 0.94 1.0 0.94 0.895 0.8 0.707 P /EoI 0.65 0.742 0.827 0.9 0.982 0.999 0.978 0.941 0.895 Cos y 0.65 0.742 0.827 0.9 0.982 0.999 0.978 0.941 0.895 Fff Ei tin. = p~" ' 0.834 0.864 0.89 0.903 0.907 0.906 0.89 0.865 0.835 App. eff. = Ei 0.54 0.64 0.735 0.813 0.894 0.9 0.87 0.813 0.74 Eol Regulation = Eo - E 0.31 0.25 0.21 0.16 0.12 0.08 0.03 -0.015 -0.05 E Problem 51. Draw vector diagrams for the cases of i' 1, 0.5, 0, 0.5, 1, of problem 50 showing the relative positions of E , E and /. Also plot the curves of regulation vs. power factor of generator and load. The preceding example, like many others included in this book, is con- structed on the basis of percentages. That is, by choosing e = 1 and i = 1, whence p = ei = 1, the results obtained may be made to apply to any case in which the constants, r and x, give the same percentage of ri and xi 0.1 = 10 per cent., ~ = 0.2 = 20 per cent. drops. In this example, If, now, 10 per cent, resistance drop and 20 per cent, reactance drop be specified, let it be required to find, with varying power factor of the load, the same quantities determined in problem 50, when v e = 2200 volts and i = 300 amp. All that is necessary, now, is to multiply those quantities representing voltage by 2200, those representing current by 300, and those representing watts, or volt-amperes, by 2200 X 300 = 660,000. METHOD IN TRANSMISSION LINE CALCULATION 119 Thus, for i' = - 1 X 300 = - 300, E = 1.31 X 2200 = 2882, / = 1.41 X 300 = 423, E I = 1.85 X 660,000 = 1,221,000, P = 1.2 X 660,- 000 = 792,000. The other quantities sought power factor of load and of generator, efficiency, apparent efficiency and regulation are the same as already calculated. The advantages of problems on the percentage basis are thus quite obvious. Problem 62. A transmission line 1 mile long supplies power to a load of 100 kw. and 1000 volts at power factor of 0.8 and frequency of 60 cycles. The line is composed of two parallel No. 000 B. & S. wires, 18 in. apart. Find generator voltage, current, power factor, power output, line effi- ciency, apparent efficiency, regulation, with the current both lagging and leading. The resistance of No. 000 B. & S. hard-drawn copper wire may be taken as 0.06 ohm per 1000 ft. at 20C. The reactance is 2wfL, where L is the inductance, in henrys, per centi- meter length of wire. L may be calculated from the formula, 10 9 where D and r are, respectively, the distance between centers of wires and the radius of the wire (Fig. 95). D FIG. 95 CHAPTER XVII CONSTANT POTENTIAL-CONSTANT CURRENT TRANSFORMATION It is sometimes desirable that the current in a circuit shall remain constant while the load varies. In series lighting circuits, for example, the current through each lamp must be nearly constant, while the number of lamps may vary from none at all up to the most that the system can sustain. Generally, however, it is desirable that the energy shall be supplied from a source of constant potential, such as a constant potential generator. Such a system is possible with a circuit arrangement like that shown in Fig. 96. Here, a high resistance, r, is placed in series with the lamps. When the lamps are comparatively few, changing their number will not alter the total resistance of the circuit very much, and the current will therefore be fairly constant. This arrangement is not, however, economical, as a large pro- portion of the power developed is always lost in the resistance, r. FIG. 96. FIG. 97. We may, therefore, try substituting inductive reactance, x, for r, and determine if this will give better results. In this case, Fig. 97, let the generator voltage be unity, that is e = 1. Let the largest value of the permissible current, in the circuit also be unity, that is, / = 1, and let the resistance of the lamps, that is, the number of the lamps, vary. We may then find the maximum resistance of the lamps which may be obtained without reducing the current lower than, say, 0.925, which will be considered the minimum, permissible current. The current is obviously a maximum when the resistance is zero, that is, when no lamps are used. e I Then, x = j = j = 1 ohm. 120 CONSTANT CURRENT TRANSFORMATION 121 Let r be the resistance of the lamps, r is variable, depending on the number of lamps in circuit at any time. Let E be made the zero vector, = e e e e (r - jx) Then where I = 7 = e Tabulating for varying r: (38) r 0.1 0.2 0.4 0.6 0.8 1 r 2 + X 2 . 1 1 01 1 04 1 16 1 36 1 64 2 a 0.099 0.192 0.345 0.441 0.488 0.5 b -1 -0 99 -0 96 861 -0 735 -0 61 -0 5 a 2 o 0098 0369 119 195 238 25 6 2 1 0.98 0.921 0.741 0.54 0.373 0.25 a 2 + 6 2 .. 1 9898 9579 86 735 611 0.5 Va 2 + 6 2 7 1 1 0.994 0.994 0.978 0.978 0.927 0.927 0.857 0.857 0.782 0.782 0.707 0.707 It is evident from the calculation that the limit of current is reached by a resistance of 0.4 ohm. This resistance could evidently be obtained by directly substituting the value / = 0.925 into Eq. (38) and solving for r. However, it is frequently prefer- able to carry out the tabulation, thus gaining the material for plotting the curves. These curves are far more instructive than the mere numerical answer. In this case, where reactance has been used instead of resist- ance in order to obtain (approx.) constant current, all the energy is consumed in the lamps themselves since the reactance (assuming zero resistance) does not consume any energy. Thus the system is efficient. However, the power factor appears to be very low. Problem 63. Determine the power factor of the circuit and plot it against the resistance of the lamps. Altogether, it may be said that in practice this arrangement is cheap and practical. A constant-current "tub" transformer has a higher power factor, but is also more expensive. In this machine regulation is obtained by altering the reactance in the circuit by means of the repulsion between the primary and the secondary coils. Problem 64. A constant-current system is supplied with power by a 122 ELECTRICAL ENGINEERING generator at 2300 volts and 60 cycles. The line resistance is negligible. Each lamp has 6 ohms resistance. Current must be maintained between the limits of 7.2 and 6 amp. Find the maximum number of lamps, both by the "resistance" and by the "reactance" method of obtaining constant current. Plot and compare curves of number of lamps vs. current for the 2 cases. Solution. (a) Resistance method. Let r = res. in series, TL = res. of lamps. _ 2300 Then r = -=-^ r + -TL = r L = No. lamps = 2300 = 320 ohms. = 383.3 ohms. 63.3 ohms. 383.3 - 320 ~ = 10.5 = 10 lamps. Lamps 2 4 6 8 10 12 TL 12 24 36 48 60 72 r + r L .. 320.0 332.0 344.0 356.0 368.0 380.0 392.0 I 7.2 6.93 6.68 6.46 6.25 6.05 5.87 (6) Reactance method. Neglecting the resistance of the reactance coil, 2300 rL 2 + 320 2 No. lamps 2300 6 383.3 210.8 = 320 ohms. = 383.3 ohms. /. T L = 210.8 35.1 = 35 lamps. Lamps 10 20 30 35 40 T L r, 2 . A 60 QfiOO 120 14 400 180 Q9 CJOO 210 44 200 240 K7 700 z 2 - 102,500 102,500 102,500 102,500 102,500 102,500 r L *+x* 102,500 106,100 116,900 135,000 146,700 160,200 W+z 2 / 320 7.2 326 7.05 342 6.72 369 6.23 383 6.00 400 5.75 NOTE. In an example of this kind it may be as convenient to work directly, without the use of complex quantities as has just been done. With more complicated circuits, however, it may be far more convenient and far safer to adhere strictly to the complex method. There are many other schemes for obtaining constant current. Some of these involve the use of condensers which are treated in the next chapter. This subject will be discussed further in Chap. XXI. CHAPTER XVIII CAPACITY AND CAPACITY REACTANCE Two conducting surfaces, insulated from each other, are said to possess electro-static capacity. Such an arrangement em- bodied as a piece of apparatus is called an electrical condenser. Condenser. When the plates of a condenser are connected respectively to the positive and negative terminals of a direct- current generator, the condenser becomes charged. That is, when a switch, s, Fig. 98, is closed, completing the circuit con- taining the generator and the condenser, ammeters A, placed in the leads, will indicate a momentary current in the direction of the arrows. No current, in the ordinary sense, could pass between the t plates. The phenomenon thus resembles the piling up of electricity, as so much ma- F terial, on one plate, the positive plate, since it is connected to the positive terminal of the generator, and the withdrawing of an equal amount of electricity from the other, the negative plate. This quantity of electricity which seems to have been transferred from one plate to the other is the charge placed upon the condenser. The condenser is maintained in an unstable state by the e.m.f. of the generator. If the generator is disconnected, the condenser continues to remain charged so long as its plates remain insulated from each other, but as soon as electrical connection is made between them, the condenser dis- charges itself by a rush of electricity from the positive to the negative plate, as indicated by the flow of electricity or current through the meters. If the condenser is charged to a difference of potential which is excessive, the insulating dielectric breaks down, allowing a discharge to take place between the plates. This indicates that the dielectric is placed under a strain when the condenser is charged. In fact, the dielectric behaves much like an elastic medium compressed between plates. When the pressure is removed the medium assumes its normal condition. The plates act merely as carriers or distributors of the charge, 123 124 ELECTRICAL ENGINEERING while its actual seat, as found out by FRANKLIN, is the surface of the dielectric. The capacity of any given condenser is determined by the dimensions of its plates, their distance apart, and the nature of the dielectric which separates them. Capacity is not a property solely of apparatus arranged in the form of a condenser, but any body may be said to possess capacity for instance, a metallic sphere, insulated and isolated in space. But this may also be considered as a limiting form of condenser in which one plate is the surface of the sphere and the other is a surrounding sphere of infinite radius. In this case the strain in the dielectric may be represented by the lines of force, or tubes of force, extending radially outward from the surface of the sphere and terminating on the surface of the imaginary sphere infinitely distant. This conception of lines, or tubes of force, due to FARADAY, makes the direction of a line or axis of a tube the direction of the force at any point, and the number per square centimeter, or density, of lines or tubes be- comes a measure of the force at the point. FARADAY assumed that the number of tubes is the same numerically as the charge per unit surface, and that the number of lines emanating from a charge Q is $ = 4?rQ. Thus each tube contains 4ir lines of force. By connecting a sphere to one terminal of a battery, Fig. 99, and connecting the other terminal to earth, assumed infinitely distant, we establish a number of electric lines / of force extending outward from the sphere. The number of lines established is 471-Q, 1 where Q is the amount of the charge placed upon the sphere. If the sphere is in air, the practical limit to ^FIG 99 ^ ne num ^ er f nnes which it is possible to establish is very closely 100 per sq. cm. of sur- face. Thus, to produce 100 lines per sq. cm. requires a charge 100 Q = -7 = 8 absolute electro-static units per sq. cm. To increase the number of lines established in any given case, the difference of potential, or voltage, should be increased. These lines are conceived to displace the ether, until by continu- ally increasing the voltage, the crowding of them becomes so 1 See Advanced Course in Electrical Engineering. \\ I / " CAPACITY AND CAPACITY REACTANCE 125 great that the dielectric breaks down. The ability of any dielec- tric to withstand rupture under the strain of potential-difference is called "dielectric strength." With a parallel plate condenser, Fig. -f~ 100, the lines or tubes are parallel, zip except at the edges, where they bow z =jT outward. By definition, the* charge due to current i during interval dt is: dq = idt. (39) The practical unit of charge or quantity, q, is the coulomb. Another fundamental relation is that q = Ce (40) where C is the capacity, and e the e.m.f. or difference of potential. This law, found experimentally, shows that the number of tubes which can be set up in a condenser of capacity C depends directly on the potential difference. In practical units, the charge in coulombs is equal to the product of the capacity in farads and the potential difference in volts. "Charge" is not a material quan- tity, but may well be thought of as a measure of "tubes." Substituting from (39) into (40), since *-*, and dq = Cde, which is called the charging current, or capacity current of the condenser. Assuming a sine wave of e.m.f., impressed on a condenser then, e = E M sin wt (42) i = CE M w cos orf. (43) The capacity, C, is a constant of the circuit, that is, like resist- ance and inductance, it is a quantity fixed by the mechanical arrangement of the circuit. Eq. (42) may be written: i = CE m o> sin (cd + 90). (43a) 126 ELECTRICAL ENGINEERING Comparing (42) and (43a) it is seen that the charging current is 90 in time phase ahead of e. Also, The effective value of the charging current is then whence E X c . / 27T/C The quantity x c is called capacity reactance, and its use in cir- cuit calculations is similar to inductive reactance. Expression of Condensive Impedance. It has been shown that the charging current leads the impressed e.m.f. 90 in time. Thus, if the charging current / is made zero vector, the im- pressed e.m.f. is jkl where k is some constant and is obviously x c . Thus E = jxj. In an inductive circuit the current lags 90 behind the im- pressed e.m.f. Thus E = jxl. Convention has settled that an inductive impedance is Z = r + jx\ thus the condensive impedance is Z = r jx c where x c as well as x is always a positive number. Circuit Containing Resistance, Inductance and Capacity in Series. To find the current. Let E, the impressed e.m.f., be the zero vector. Then / = - -A- - = r + jx - jx c where a = b = - ^X X c ) (x - x c ) r (x - z c )- To find the voltage E c across the condenser. We have: EC = I(-jXc) = e(a H- jb)( jx c ) = e( ajx c + bx c ). Similarly, the voltage across the inductance is E L = I X jx = e(a + jb)jx = e(ajx bx). CAPACITY AND CAPACITY REACTANCE 127 Problem 56. Let the constants of a circuit be r = 1 ohm, L = 0.0265 henry, C = 0.000265 farad, and let 100 volts be impressed on the circuit at variable frequency. Find, and plot against the frequency, 7, E c , EL, E r for frequencies from to 100 cycles per sec. Jl /. i fo Solution. We have: 7 *np- f c e(a + jb); I = eV a 2 + b 2 . E c = e( - ajx c + bx c ); E c = E L = e(ajx - bx); E L = Also, E T = e(a + j'6)r; E r = er\/a 2 Tabulating: FIG. 102. / 20 40 50 55 60 65 70 100 2irf 125.6 251.2 314.0 345.2 376.8 408.0 440.0 628.0 X 3.33 5.65 8.33 9.15*' 10.0 10.8 11.65 16.66 x c 30.0 17.65 12.0 10.92 10.0 9.25 8.57 6.0 (x - x c ) - CO -26.67 -12.0 -3.67 -1.77 0.0 1.55 3.08 10.66 (X - X C )2. . . . + oo2 712.0 144.0 13.5 3.14 0.0 2.4 9.5 114.0 r2 + ( x -x c )2.. oo2 713.0 145.0 14.5 4.14 1.0 3.4 10.5 115.0 a 0.0014 0.0069 0.069 0.242 1.0 0.294 0.095 0.0087 b 0.0374 0.0827 0.253 0.428 0.0 L0.455 -0.293 -0.0925 a* 0.000002 . 000048 0.0048 0.059 1.0 0.086 0.009 0.000076 62 0.0014 0.0068 0.064 0.183 0.0 0.207 0.086 . 0086 (I* + 62 0.0014 0.0069 0.0688 0.242 1.0 0.293 0.095 . 0087 Va 2 + 62. ... 0.0374 0.0828 0.262 0.492 1.0 0.54 0.308 0.093 7 3.74 8.28 26.2 49.2 100.0 54.0 30.8 9.3 EC 112.2 146.1 314.0 537.0 1000.0 500.0 264.0 55.8 EL 12.5 46.8 218.0 450.0 1000 . 583.0 370.0 155.0 E r 3.74 8.28 26.2 49.2 100.0 54.0 30.8 9.3 40 60 Frequency FIG. 103. 128 ELECTRICAL ENGINEERING Resonance. Curves of the form shown in Fig. 103 are called resonance curves, and their maximum points of the de- pendent variables are called resonance points. In this case, it is said that 60 cycles is the frequency of resonance. On examining the problem it is seen that resonance is attained at that frequency for which x x c = 0, or when the effect of inductance is just nullified by that of capacity. The circuit then behaves as though it possessed resistance only. CHAPTER XIX PARALLEL CIRCUITS Let 1 1 and 1 2 be any currents in the branches of a parallel system, such that I\ = i\ + ji f \ and 7 2 = it + ji f z- Laying off these vectors (Fig. 105), and adding them, gives 7 = 7i + 7 a = ii + la + j(i'i + *' 2 ). (44) Let the impedances of the branches (Fig. 104) be Zi = ri jxi and Z 2 = r 2 + j^2, respectively. FIG. 104. FIG. 105. To find the currents 7i, 7 2 , and 7. We have: . _ e where g\ = r\ + T + xi 2 is the conductance, is the susceptance, and FI = gi + jbi is the admittance of the first branch circuit. eg i is the power component of I\. cbi is the wattless component of I\. Similarly, _. ^2 JXz) , . ., v _ y where _^^ r J and r 2 /= (45) 129 130 ELECTRICAL ENGINEERING To find the joint impedance, Z, of the branches, e_ _e_ !_ ~~ I ' eY '' = Y' Example. In Fig. 104 let n = r z = 0, and Xi = ^- f ^> x- 2 = 2ir/L. Then - = 0, 1 " z 2 27T/L' Then, from (45), 7 = e (o + j(2*/C - g^) ) (46) From this it is seen that the line current is in time quadrature with the voltage. If I = 0, then from (46) we have the relation or = 1. that is to say, that if in a circuit such as is here considered the frequency be varied, a value may be reached for which the line current will be reduced to zero. In such a case the currents in the branches will be 7i = e(0 Both of these currents are in time quadrature with the voltage, but /i is leading while 7 2 is lagging. Thus, they are in time phase opposition to each other. Problem 66. In the circuit of Fig. 104 let e = 100, n = r 2 = 1, L = 0.0265, C = 0.000265. Let the frequency vary, as in problem 54. Find / Ii, Jzj and plot them against the frequency. Transmission Line Supplying Power to Parallel Loads. Let a transmission line of impedance Z Q = r Q + j%o be used to supply power to a load consisting of two impedances, Zi = 7*1 + jx\ and 2 = r 2 + jxz, which are in parallel. Besides the impedances, let E the voltage at the receiving end be known. Find 7, /i, 7 2 , E Q , P.F. of generator and of combined load, regulation and efficiency of the line. E is chosen as the zero vector = e. PARALLEL CIRCUITS 131 Then where Zi ri+jxi rf + xi' (0i + jbi),. Similarly, where /2 = FIG. 106. And where Then / = /I + / 2 = m + jn, = e (0i 4- 02 + j(&i + 62) \ __ _/t i^ t \ = e + (w -f jn)(r + jx ) nx Q + j( nr o - Eo = e + = e + = a where a Q = e -{- mr nx 0) 6 = nr Power of generator, by telescoping E Q and I =P Q = a m + 6 n. PO a w -f- 6 ?i .'. P.F. of generator = P.F. of combined load = -v = F' EJ - e Regulation = , . P em Efficiency = ^- = Problem 67. In the same circuit (Fig. 106), let E be known and E unknown. Find all the quantities obtained in the last problem. NOTE. In solving this problem the student is again urged to pay particular attention to the form of his work. In order to add emphasis to this matter these similar problems are here given, the one being worked out and the other left for the student to do. 132 ELECTRICAL ENGINEERING The numerical or scalar expressions are not put down. It is assumed that they may always be obtained when needed by the simple process of rationalizing a simple complex expression. By omitting them in the process, confusion is eliminated. Approximate Transmission Line Calculation. The two parallel wires of a transmission line may be regarded as constituting the plates of a condenser. When alternating e.m.f. is impressed upon the line there will therefore flow a charging or capacity current over the line, whether the distant end is open or closed. Fig. 107 gives an approximate representation of such a line in ___ ' FIG. 107. which the line capacity is replaced by two condensers, one at each end, so proportioned that each shall take one-half of the charging current. The charging current is taken as 2i 2 . ii is always positive, whereas i', the wattless component of the load current, is positive or negative depending on the load. Then EQ = e + I Z = o + j where ao = e + ir Q i s x Q) /o = / + jit = i + j(i' + 2i 2 ) = i + ju. From these, the power, power factor, efficiency, etc., may be determined. Expressions should be obtained by the student for practice, as follows: Power given by generator = P = Apparent power at generator = E Q I Q = p P.F. at generator = cos <*o = jrj~ Efficiency of transmission = ,5- = * o Apparent efficiency CHAPTER XX DISTORTED WAVES. RESONANCE EFFECTS So far, only current and voltage waves have been dealt with which followed a sinusoidal variation with respect to the time and had the same frequency or period. In the laboratory, re- sults obtained are found not always to agree with those expected from the theory. This is frequently due to the assumption in theory of pure sine waves, whereas, in practice, a pure sine wave is only approximately attainable, and the actual waves may differ greatly from that form. It can be proven that any curve representing changes occurring with time can be resolved into a number of sine waves of differ- ent frequency as long as the curve representing the changes is a univalent function of time which it always is in electrical problems. It can also be proven that if the curves traced are symmetrical above and below the axis no matter how distorted the sine waves contain only the odd frequencies. Thus assume as the simplest case that the current is distorted in such a way that it can be represented by the first two terms of the series, that is that: i = Ii m sin wt -f J 3m sin a). It is seen that the frequency of the second component wave is three times that of the first. The first wave is called the fundamental of the complex wave, the second wave is called the third harmonic. The angle a denotes the per- manent phase difference be- tween the waves. Such a combination of waves is seen ina rIG. luo. to be a distorted wave, as shown in Fig. 108. To find the amount of heat such a wave will develop in a circuit, that is, to find the effective value of the complex wave. 133 134 ELECTRICAL ENGINEERING Evidently the heat developed at any instant is proportional to i 2 , and i 2 = [Iim sin ut + 7 3m sin (3co -f a)] 2 . The mean value of the heat developed during a cycle of the wave will then be proportional to 1 C T mean i 2 = I [I lm sin ut + h m sin (3co + a)] 2 dt. (47) Thus, the effective value of the current is \/mean^ 2 = 7 = \/ I t^im sin cot + 7 3w (3co + )] 2 cfr. (48) The student should solve (47) and (48), and show that 1 2 7 2 . -Mm , -* 3m mean z 2 T^+jT (49) and / = \/Ii 2 + /a 2 where 7 im = maximum value of the fundamental current wave, 7 3m = maximum value of the third harmonic, 7i = -^7= = effective value of the fundamental, 7am 7s = j= effective value of the third harmonic. V2 Also, in general, where there are any number of component harmonic waves in a circuit, / = Vli 2 + h 2 + 7 6 2 + . . . (50) Thus is found the important rule that the effective value (ammeter reading) of any number of currents of different fre- quencies is equal to the square root w of the sum of the squares of the in- dividual effective values. 15 NOTE. Eq. (50) holds for any combina- tion of harmonics whatsoever. With alter- nating-current machinery, we have to deal only with odd harmonics, as the positive and negative waves are always symmetrical except during transient periods not considered in this volume. Example. In Fig. 109 are represented three generators which supply respectively 20 amp. at 60 cycles, 15 amp. at 25 cycles, and 10 amp. at 10 cycles. They all use a common wire for a part DISTORTED WAVES. RESONANCE EFFECTS 135 of their circuits. Then the current which flows in the common wire is / = V20 2 + 15 2 + I6 2 = 27 amp. Problem 68. If still another generator is added to the above system and it supplies 12 amp., direct current, to its load, using the common wire, find the current, I, that will then be in the common wire and explain the result. E.m.f. Which Causes Distorted Waves of Current. If the current in any circuit is given by the equation i = 1 1 sin at + 7 3 sin (3orf + a) (51)' the question may naturally arise as to what kind of e.m.f. wave will cause such a current to flow. Will the e.m.f. wave be more or less distorted when the current is supplied to a circuit of re- sistance and inductive reactance? We have (Eq. 15), , T di e = ir+L dt' Substituting from (51), e = Ijr sin wt + 7 3 r sin (3co + a) + L(/io> cos co + 37 3 co cos (3co + a)) = 7 if sin ut + L/io) cos coZ + 7 3 r sin (3o>Z + a) + 3L7 3 o> cos (3coZ -f a) lir sin ut + IiX cos ut + I s r sin (3orf + a) + I 3 x 3 cos (3orf + a). (52) Let - = tan /?; = tan /3 3 . Then r = Zi cos j8 = 2 3 cos /3 3 . Substituting these values in (52), e = /ii(cos j8 sin co + sin cos $)+ /3^s(cos 3 sin [3orf + a] + sin & cos [3orf + a]) = JiZi sin (coi + 0) + 7 3 Z a sin (3w< + a + ft) = Ei sin (co + 0) + Ei sin (3arf + + ft)- ( 53 ) Thus the amplitude, EI, of the fundamental voltage wave is Zi times that of the current fundamental; the amplitude E 9 of the triple frequency voltage wave is Z 3 times that of the cur- rent triple frequency harmonic. 136 ELECTRICAL ENGINEERING The difference between the multipliers, Zi and Z 3 , is due to their respective reactances, x and x 3 , since r is the same in each. But z 3 is 3x. Therefore, it is seen that the triple frequency voltage wave is greater in proportion to its fundamental than the triple fre- quency current wave to its fundamental. In other words, the voltage wave is more distorted. Conversely, it may be said that when a distorted voltage is impressed on a circuit, the effect of the inductive reactance is to smooth out some of the distortion in the current wave. Problem 69. Show that when the e.m.f., e = EI sin (at + E 3 sin (at + a), is impressed on a circuit of resistance- only, the current flowing will have the same amount of distortion as the voltage has. Problem 60. Show that when the e.m.f. of problem 59 is impressed on a circuit containing resistance and capacity, the effect of the capacity is to increase the distortion of the current. If the voltage (53) is measured by a voltmeter, what will the reading be? From the development of (51) in respect to dis- torted currents, since both currents and voltages are similar in form it follows that the effective e.m.f. shown by a voltmeter will be E = Problem 61. In Fig. 110 let E be the known impressed voltage, let capacity = C farads, inductance = L henrys and resistance = r ohms. Then X e = ; XL = 27T/L. Find the current, and the voltage drops across the inductive impedance and the capacity, when the impressed voltage is composed of a fundamental and a third harmonic. The fundamental component of current will be where b = - Zo" and Zo 2 = r 2 + (XL x c )*, and EI = e\ is the zero vector. DISTORTED WAVES. RESONANCE EFFECTS 137 The voltage drop across the inductive impedance, z, due to /i, is Eiz = I\Z = ei(g -{- jb)(r -\-JXL) = a + jb' where a = e\gr e\bx] b r = (br + gx}e\. The voltage across the capacity reactance due to /i is Eic = 7i(0 - jxc) = d +jf, where d = eibxc] f = e\gx c . The third harmonic components of current and voltage are similarly deter- mined, remembering that x 3L = SZL, _ _?5. Problem 62. In the circuit of Fig. 110, let r = 1, L = 0.0265, C = 0.000265, E = 100, E 3 = 30. Find and plot the current waves /i, 7 3 and 7, as the fundamental p IG HQ frequency is varied. NOTE. Solve for frequencies of 15, 20, 25, 35, 50, 55, 60 ,65, 75, 100. Solution. We have, first, *T Then where 7 3 = where - # 3 2 = lOO 2 - 30 2 = 95. 95 95r-a;i) 95r 95(r-.ysi) _ r 2 + zi 2 r 2 + 1 = XiL ~ 30 30r + +ji f 3, = x 3L - 1 600 200 = 2./3L - ^77, = 0.167/a - 77 = 0.5/t - -jr- 27T/ 3 C Waves of 7 lf 7 3 and / are plotted in Fig. 111. It is seen that the maximum /i occurs when XIL t * when 0.167/1 = -7-. or, at the frequency /i = 60, and it is I lm = -j = 95 amp. 200 Maximum 7 3 occurs when x 3 L = x 3 c, i-e-j when 0.5/i = -71 or, at the fre- 30 quency /i = 20 and it is I sm = ~r = 30 amp. 138 ELECTRICAL ENGINEERING CN CO "5 b- O t>.iOiCOOOOOcOiOCNOOO O OOOOOO>OC5O>O5(N CO CO O^iQOGOOOON-OOOOO W OO^iCOOrHOCOCOrHOO -> t* iQ ^OQtOi t* "* "" CN 00 -^Oi-H^ T*iOcOOO-r>-OOOCOi O Ci iHCNCOCOC.'0 iHTH ^l rH rH CNr-ICOC5 (O I-H 1C t t >C rH O5 tN.OOCOCOO5OOOl- 9 1 CO CO CN i * CO "^f ^ O COOOt^OOOOOOCOO . r-I^HCN(N t^ 1-4 r-4 O oJOoot- COt>. t-( ^00?<0000000000 i-HCOTf< -iO5 O iCCCOO3COOO3OCOO Mi-HlNIN .Ot>.O O U5 CN (N 5| iO r-i O O O I-H 00 O GO l> CN O* t^ CD t^ O CN i CN O CO CD* CN b^ ^ CO O O CD CO 00 OiC COCO "3 NCO >CM CNC CO 0000000"5t*5(N>0'*iM<0^ CO U3M, Substituting values for xi and z 2 , 125 *" 1252 (57) then becomes (58) becomes (60) becomes 250 -lM- 2r - r / = = 2 = constant for all values of r. + 12 2 62.5 This case is, of course, ideal in that it assumes absence of re- sistance in both reactive branches of the circuit. In Chap. XVII, it was found that with the system which used only inductive reactance to obtain constant current, the power factor was quite low. To obtain an expression for the power factor in the present case, h jl) = eh + jel, ,e whence, by telescoping, the power is _o n ^ m e i x\\ e x\ e = ( 1 -f- ) - =- r \ xj x 2 r r Substituting numerical values, P = 4r. CONSTANT-CURRENT TRANSFORMATION Volt-amp. = #0/1 = (eh + jel) (- - j -} \T 3/2/ e% eH 143 Substituting values for h and I, This equation reduces to when Xi = x^ and, substituting numerical values, 1 = 4Vr 2 + I25 2 ' .'. Power factor =W^F~ = Tabulating : 2 + 125 2 T o 10 20 50 100 200 500 e 20 100 40 400 100 2,500 200 10,000 400 40,000 1,000 250,000 r 2 + 125 2 . . . 15,625 15,725 16,025 18,125 25,625 55,625 265,625 Vr 2 + 125 2 P.F /i 7 2 125 2 125.2 0.080 2.00 16 126.5 0.158 2.02 32 134.5 0.372 2.15 8 160 0.625 2.56 1.6 236 0.847 3.78 3.2 515 0.97 8.25 8.0 Fig. 114 shows the curves of current, voltage, and power factor for change of resistance of the load. In this case, the power factor is seen to be very much better than it was found to be where inductive reactance alone was used. Problem 64. Let the load in the preceding problem be made up of both r and x. Find the effect of reactance in the load and make a general study of the conditions under varying power factor of the load. 144 ELECTRICAL ENGINEERING This may be done as follows: Imagine the load to consist of any number of lamps, each lamp possessing a certain resistance and a certain reactance. Then the ratio of - will be constant. 1. Let x = 0.5r. The power factor of the load will then be 1 V/r 2 +0.25r2 Vl.25 0.895 2. Let x = r. The load power factor is then 0.707. Supplying these values in turn to Eqs. 54, etc., as in the previous problem, tabulations and curves may be obtained from which a report on the effect of power-factor variation of the load may be made. In order to bring the subject to a practical basis, the effects of actual constants of the apparatus should be investigated. It 400 500 Resistance FlG. 114. will be found that the resistance of a suitable reactive coil for such a case as developed above, would not need to be above 0.5 ohm, and that the resistance in the condenser circuit would be very much less. Consequently the effects of resistance are extremely small, and the case, as worked out, may be considered as approximately attainable in practice. Many schemes have been proposed for the attainment of con- stant current from a constant potential source, in which more or less elaborate combinations of reactances have been arranged. A CONSTANT-CURRENT TRANSFORMATION 145 study of the possibilities of different schemes is profitable for the student as it affords excellent practice in circuit calculation. 1 Power and Wattless Components of Volt-amperes. The quantity El cos a is called the power component P of the volt- amperes. By a similar conception, El sin a is called the " wattless" component P 1 of the volt-amperes. Thus El = V(EI cos a) 2 + (El sin) 2 - Referring to Eq. (36) El sin a = El sin (0 - 7) = El (sin )8 cos 7 cos sin 7) = EI ^]^ = i'e - e'i. (61) 1 For some of these developments see STEINMETZ, "Alternating-current Phenomena," Chap. X. 10 CHAPTER XXII THEORY AND USE OF THE WATTMETER The most accurate way of obtaining results in the measure- ment of alternating current, voltage or power is by the use of the electro-dynamometer. As generally employed the electro-dynamometer, invented by SIEMENS is a combination of 2 coils, one movable and the other fixed, whose planes are set at right angles to each other. When current is sent through the coils, each sets up a magnetic field in the region occupied by the other, thus causing forces which tend to move the coils relatively to each other. The forces are balanced by tension on a calibrated spring. If the same current is sent through both coils, the scale, prop- erly calibrated, measures the current. If the instrument is placed in series in a circuit, the current measured is that of the circuit, and the meter becomes an ammeter. If it is placed in shunt to a given circuit, the current in the coils is proportional to the voltage drop in the circuit, and the meter becomes a volt- meter. If, however, one coil is placed in series and the other in shunt to a given circuit, the effect on the instrument is proportional to the product of the amperes and the volts at any instant, and the electro-dynamometer becomes a wattmeter and measures power. 1 For practical construction, the coil to be connected in series is made of few turns of comparatively heavy wire, and is usually the fixed coil, while the coil to be connected in shunt is made of very many turns of fine wire and is movable. Accurate results are obtained by the dynamometer because the coils, while readings are made, are always kept in the fixed rela- 1 The flux set up by 1 coil (fixed) is proportional to the current flowing in the circuit, while the flux set up by the other coil (movable) is proportional to the voltage across the circuit. But the force at any instant acting on the coils is proportional to the product of the fluxes set up by the coils, that is, the force on the coils is proportional to the product, E X /, where E = voltage across the circuit, and / = current in the circuit. 146 THEORY AND USE OF THE WATTMETER 147 tive position at right angles to each other, thus eliminating mutual flux; 1 also because no iron is used in construction, and there are no other materials which might cause variation in the results. Accuracy must be obtained, however, by the correction of certain errors in the readings. The error due to friction of the movable coil is small. The error due to changes of resistance by temperature is obviated in good instruments by the use of resis- tance which is not affected by change of temperature. When used as a wattmeter, the readings of the dynamometer must be cor- rected for error due to the manner of connection in the circuit. This correction is of great importance. Let the wattmeter be connected as in Fig. 115 (A), in which the power consumed by the impedance, Z = R + jX, is to be measured. The current coil is represented by the impedance z r + jx] the voltage coil by the impedance z\ = < 7*1 -\-jx\. The load voltage is e, which is chosen as the zero vector. (B) (c) FIG. 115. The meter should be first calibrated by direct current, so that its reading for any given direct-current power is known. Wattmeter Connections. Connection (A) is wrong, because the current coil has to carry /i, the current taken by the voltage coil, in addition to the load current /, thus causing the wattmeter to indicate the power lost in the voltage coil, plus the load. There will thus be a reading even at no-load. If JiVi, the power lost in the voltage coil, is subtracted from the wattmeter reading this error is eliminated. This error is usually negligible in connection with circuits carrying large current at low voltage. In the connection shown in Fig. 115 (B), the current coil carries only the load current. The voltage coil, however, is so connected as to include the drop in the current coil as well as that across the load. Thus the wattmeter indicates the power 1 Mutual induction will be treated more fully in connection with the study of the transformer. See Chap. XXVI. 148 ELECTRICAL ENGINEERING lost in the current coil in addition to the load. This error may be corrected by subtracting the Io 2 r power lost in the current coil. Connection, (B), is best adapted to measurements at high voltage and low current. A third arrangement, Fig. 115, (C), is known as the compen- sated wattmeter. In this there is wound a fine wire coil of the same number of turns as the current coil directly upon the latter. By its connection, it is seen that this coil, c, carries the current of the voltmeter coil. It therefore supplies to the current coil just enough back ampere-turns to neutralize those due to that excess of current in the current coil. This arrangement causes the wattmeter to read correctly so far as its connections are con- cerned. Readings of the dynamometer calibrated by direct current must also be corrected for an error due to change in frequency. In connection (A), Fig. 115, the load current is Current in the voltage coil is e Current in the current coil is /o = 7 + 1 1 = e(g Q + jb ). \ These currents are plotted in Fig. " e l!6. Usually, the angle of lag of /i, is very small, even less than 1. This slight lag may, however, give a large error. Tan 7 = Since the coils are at right angles, the torque at any instant on the movable coil is proportional to the product of the currents. Thus, T = ki i 1} where i Q and ii are instantaneous values of current in the two coils. THEORY AND USE OF THE WATTMETER 149 Let io = I 0m sin coJ, and ii = I lm sin (ut + a). Then the average value of the torque through one-half cycle is fc/ - I (/ Ow sin ut)(Ii m sin (co< -f a)dt /i A . I (si . sin 2 co< cos a + sin ut cos w sin a)dt o/] Ah - cos2o; sin2co 1 I ~~2~~ ~ COS a "^ --- 2 Sln a 2fc/o/i["fa)< cosa __ cos a sin 2co^ sin a cos 2 due to the inductance of the coil. V ri 2 + Zi 2 The reading should therefore be corrected by the factor in order to bring it proportional to e. (b) The current in the current coil is too large in the ratio, -j The correction factor is therefore, L /o 150 ELECTRICAL ENGINEERING (c) The correction factor for angular displacement is evidently ^-^i where and a are as shown in Fig. 116. The complete cos a correction factor is then 21 IG 2 + B 2 cos ft The constants of the wattmeter are assumed known, which per- mits of obtaining all angles except the phase angle of the load. Thus, a is known, but not 0. In order to obtain 0, a reading may be taken of the wattmeter, voltmeter and ammeter as in the ex- ample below. Then, roughly, = ~- (62) Substituting this value of cos into the correction factor, a new value of W is obtained. Replacing the approximate W of (62) by this new value, a new value of cos is obtained in which the error is of the second magnitude. By repeating this process any desired degree of precision may be obtained. Example. To find cos /3, when by reading of instruments the approxi- mate power factor is found to be = _ = 50 volt-amp. 100 pow^r Factor' 25 There must be a correction-factor Fm. 117. curve of the dynamometer for varying power factor. From this curve, let the value of k be 0.99 f or P.F. = 0.5. Then multiplying, 0.5 X 0.99 = 0.495 = power factor to second ap- proximation. It is evident that a repetition of the process will be hardly necessary in most practical cases. Problem 66. With the wattmeter connected as above (115, A), determine and discuss the correction factors: (1) with non-inductive load; (2) when the power factor of the load is just equal to the power factor of the voltage coil; (3) in the theoretical case when there is no self-induction in the voltage coil. Problem 66. By a process similar to that just given, find the correction factors for wattmeters when connected according to (Fig. 115, B and C). The errors actually obtaining in practice with good commercial indicating wattmeters are quite small. Thus, at normal voltage, THEORY AND USE OF THE WATTMETER 151 2000 cycles and power factor from 0.8 leading to 0.8 lagging, the error is usually less than J4 per cent. As the power factor is lowered the error becomes larger. At normal voltage, 60 cycles, the error may be less than 0.2 per cent, with the power factor down to 0.1. If the impressed voltage is low, say 15 per cent, of normal, the error may, however, be several per cent. In operation there are also errors which enter with the use of "current" and "potential" trans- j| g formers. When these are used, the error is |^iovoit 5 practically negligible for power factors above f ' _j 0.8 except for small loads. With non-inductive- load of, say 10 per cent, normal, the error may, however, be several per cent. Problem67. An uncompensated wattmeter (Fig. 115, A &ndB and Fig. 118) has a rating of 400 watts. At 100 volts the resistance of its voltage coil is 2000 ohms. At 50 volts the resistance is 1000 ohms, at 10 volts it is 200 ohms. The inductance of the voltage coil is 0.007 henry. Resistance of the current coil is 0.03 ohm. Inductance of the current coil is 0.0003 henry. Find the wattmeter reading, the actual watts and the correction factor for all combinations of voltage, current and power factor, when e = 100, 50.0, and 10.0, volts 7 = 4, and 0.4 amp., P.F. = 1, 0.1 lead, and 0.1 lag. / =60 cycles. CHAPTER XXIII SIMPLE PROBLEMS IN ELECTRO -STATICS It is desirable at this point to introduce certain principles of electro-statics. These should, of course, be more or less familiar to every student who has had an adequate course in physics. Potential. By definition, the potential at a point in an electric field is equal to the work done per unit charge in bringing a positive charge from a place of zero potential (usually infinity) to the point. Intensity. Also by definition, the intensity of the electric field (lines per square centimeter in air) is numerically the same as the force which that field exerts on unit charge. FIG. 119. Thus, if R is the intensity of the field at a distance r from a point charge, Q (Fig. 119). See also Chap. XVIII. R = _ _ = = Q area of sphere of radius r 4.irr 2 r 2 Therefore the potential at p is VP = - fRdr = - fR cos 6ds (63) where ds is an element of the path of the unit charge and dr ds cos 6. The minus sign is used because work is done in bringing unit positive charge against the charge Q which is also assumed posi- tive. Thus, the repulsion between the charges must be overcome 152 SIMPLE PROBLEMS IN ELECTRO-STATICS 153 and work has to be supplied. This designation is, of course, a matter of convention. Substituting the value of R, just obtained, V P = - 1 Rdr = - \ Qdr = J- Jj 2 r_ p Q P Q where p is the distance from Q to p. Capacity of a Sphere. Suppose the charge, Q, to be on an iso- lated sphere of radius, n. Then, at the surface of the sphere p = ri, and the potential is Vi = The capacity of a condenser is defined as the charge per unit potential. Thus, C = y, where C is capacity, and V is the potential of the charge Q. Since, therefore, with an isolated sphere, Vi = i the capacity of the sphere is in cm. Thus, the capacity of a sphere is numerically equal to its radius; the value of the capacity expressed in farads, C is found by dividing C in centimeters by the constant 9 X 10 11 . Potential Gradient. The potential gradient, usually denoted by G, or the rate at which the potential changes at a given point, is of very great practical importance since it is a measure of the electric stress to which the dielectric is subjected. The potential gradient, G, and the electric field intensity, R, are the same nu- merically. Thus, if the potential of a certain point falls at the rate of 5 units of potential per cm., the actual number of lines per sq. cm. at the point is also 5. By definition, Since, dV = - Rdr . In a dielectric of specific inductive capacity, K, the intensity as well as the potential gradient for a given charge is less than in 154 ELECTRICAL ENGINEERING air. It is - times the intensity in air. Thus, in the case of a sphere, c v l Q G== R== - -,- The maximum possible value of G, or R, under ordinary con- ditions in air, is not known exactly, but is in the neighborhood of 30,000 volts per cm., or 100 electro-static units of potential. Capacity of a Spherical Concentric Con- denser. Consider 2 spherical concentric bodies with charges plus and minus Q. By (63), the potential difference is r* = r = - Rdx, Jx = n where r and n are ^radii respectively of the inner and outer sur- faces of the condenser. But Q -- r t/ n Therefore the capacity of the condenser is C = Q -?^-in cm. (65) ri -f- r Potential gradient between concentric spheres. Since rfF G = T-> dr and dV = - Rdr, But SIMPLE PROBLEMS IN ELECTRO-STATICS 155 rr c = r At the surface of the smaller sphere, x = r, whence the gra- dient is 7*1 V = 7 (r, - r)' The Capacity of a Concentric Cylinder. Let the charges be Q per cm. of length of the cylinder (Fig. 121). Then, by GAUSS' theorem, the flux emanating from each centimeter of length = 4irQ. Lines of flux are here assumed to extend radi- ally, which they actually do. At any distance, x } from the center of the cylinder the intensity at a point is the total number of lines divided by the area, or, Thus, the potential difference is: dx + 2Q log r c^Q I Rdx = I dx = 2Q[log r log rj Jn -Jri (66) and the capacity is n C = = in cm. per cm. length of the concentric cylinder. The gradient at any distance, x, from the center is x x x " _ . r, . r, (68) 2 log - x log - At the surface of the inner conductor, x = r. a r log - & r i and this is the greatest value of the gradient. 1 See "Advanced Course in Electrical Engineering." 156 ELECTRICAL ENGINEERING In these formulae no account is taken of any effects due to the ends of the concentric cylinder. For the special case of an outer cylinder of radius ri = , C = 0. Capacity of Two Parallel Plates so Large that the Effects of Their Edges may be Neglected. The total flux set up by a charge, Q, is IwQ (Fig. 122). 4x0 The intensity, R j-> where A is the area of one side of the plate. The potential difference is: A t +Q 1 f / ^ 1 1 X 1 ^ -Q 1 V JS FIG 122. -$> *> ' - where d is the distance between the plates. The capacity Q A K A C = ~~ = ~ 4 j = - A ; in cm. e 4ird 4?ra (69) (70) where the dielectric has a specific capacity K. The potential gradient, G y is a constant in the dielectric be- tween the plates, since the flux lines are parallel. Thus, C - ^Q _ ^TrCe _ efc ~ dx = ~~A A = ~A in which e is the difference of potential of the plates and k is a constant, = 4irC. Capacity of a Transmission Line. 1 The line is represented in section in Fig. 123, with, r, the radius, and, D, the distance be- tween centers, of the wires A and B. Let A be charged + Q, and"*" B } Q. The flux lines emanat- ing from A enter B. The inten- sity at a point, p, due to the charge on A, is R A ; that due to the charge on B is R B . *|^r I FIG. 123 1 For more exact deduction see "Advanced Course in Electrical Engineering." SIMPLE PROBLEMS IN ELECTRO-STATICS 157 Then r> --" v& ^vl RA - ^z = p " 2ir(D-x) ~~ (D-x) The intensity due to the two charges is the sum of R A and R B , since the direction of the lines of electro-static force from A , due to a positive charge, is the same as that due to B, which has a negative charge. The potential difference is: C r C T /I 1 \ dx e = - \ Rdx = - 2Q I (- + ~ -) JD-T JD-T \* D ~ x/ = 4Q log ^^ (71) and the capacity is therefore 1 (72) per cm. length of circuit not of wire. This capacity is expressed in centimeters. If the line is in a dielectric of specific inductive capacity, AC, the capacity in air as determined above, must be multiplied by K. To transform capacity, expressed in electro-static units in (72), into electromagnetic units, the former should be multiplied by 2 where v is the velocity of light = 3 X 10 10 cm. per sec. The practical electromagnetic unit of capacity is the farad. C Capacity in farads = -^ X 10 9 , where C is capacity expressed in electro-static units. .'. Farads = electro-static units X ..Q// 9 \/ Thus, C/cm. of circuit, in farads, = k 4 log ^~ X 9 X 10" (4 logio ^f- r ) X 9 X 10" 158 ELECTRICAL ENGINEERING When connected to a source of alternating e.m.f., the effective value of the charging current is I c = 2ir{CE, where E is the effect- ive value of the line voltage. The voltage is frequently taken from one ~E~ n " side of the line to neutral, that is, to the E --- * -------- point of zero potential of the system (Fig. | _ 124). When this voltage to neutral is used, FIG. 124. the capacity to ground, or to neutral, is twice as great as the capacity between lines. This follows since I c = 2irfC n E nj where C n and E n are capacity E and voltage to neutral, and for single phase systems, E n = ^r- z For three-phase systems, E k 2 X 9 X 10 11 X log _ , farads per cm. of line, since in r using the neutral, the length of line is the transmission distance. 0074 Reducing values to practical units, C n /1000' = - ^ _ logic is the capacity to neutral per 1000 ft. of line, in micro-farads. 7 C /1000' = 1Q n 6 is the charging current per 1000 ft. of line, in amperes. Capacity of a Three-phase Cable. Capacity to neutral per 1000 ft. of line is given in micro-farads by the formula 0.0074 i Cn/1000' V3a R* - a 2 logio r R* + a 4 + Such a cable is represented in section in Fig. 125, where R is the radius of the sur- rounding sheath, a is the distance from the center, or neutral point, to the center of one FlG - 125> of the wires and r is the radius of 1 wire. Problem 68. (a) Prove that the greatest charge which may be put on a ball of 10 cm. radius is 10,000 electro-static units. (Assume that the maxi- 1 This will be understood from later discussion of polyphase systems and deduced in the volume dealing with advanced electrical engineering. SIMPLE PROBLEMS IN ELECTRO-STATICS 159 mum gradient is 30,000 volts per cm. when air at atmospheric pressure "breaks down" and a glow called corona appears around the wire.) (6) Prove that the greatest surface charge, in coulombs per sq. cm., is ~ffp' (c) Show that if the inside conductor of a concentric cable has a radius of 1 cm., and the outside conductor is 2 cm. in radius, 0.0027 coulombs must be put into 1 mile of cable to cause it to glow (corona). Show that the potential difference between the 2 conductors is 20,800 volts. Inductance of a Concentric Cable. The inductance is recol- lected to be the interlinkages of the flux and turns per unit current. In general, if the m.m.f . acting in a circuit is F then the flux 4iTrF X area of magnetic circuit length of magnetic circuit The interlinkage factor is the fraction of the total current en- closed by the flux, and is L = 7 S flux X interlinkage factor. (73) Consider first the flux in the inside conductor due to the as- sumed uniform distribution of the current in it. At a distance x from the center (Fig. 126), TTX 2 the m.m.f. is ^ / where I is the total current. ?rr 2 The area enclosing the flux per centimeter length of conductor is dx and the length of the magnetic circuit is 2irx x* ,dx_ _ x_ . Wi = 47T r , I 2wx - L r , a TTX 2 This flux interlinks with ; of the total current; thus the FIG. 126. ?rr u interlinkage factor is 5- FIG. 127. (Assuming that /* = 1) (74) d Xl Between the conductors, the flux inter- links with the whole current (Fig. 127). Thus by a similar reasoning we get : fr = 21 '*f 160 ELECTRICAL ENGINEERING The current in the inner conductor interlinks with the entire flux which is in the outer conductor but which is caused by the difference in m.m.f. in the inner and outer conductors. At distance x the m.m.f. is thus __ " R Q 2 - R 2 R, 2 - R 2 The interlinkage of this flux with the current in the inner con- ductor is, of course, unity, thus The inductance of the outer conductor should be added to give the total inductance of the cable. The m.m.f. is shown above to be j. R<> 2 - x<> 2 R, 2 - R 2 (Ro 2 - x, 2 ) (Ro 2 - R 2 ) 2 2 ' 1 R 2 + R 2 2Ro 2 R 2 Ro 2 R Q 2 - R 2 + (#o 2 - ^R 2 ) 2 g R The total inductance L = LI + L 2 + L 3 + L 4 which is readily proven to be 1 01 R , 2/V Ro 1 3^o 2 - R* L = - 2 + 21og 7 + ^g^kg - - - R ^ _ R2 cm. This inductance is expressed in the absolute system of units. By dividing by 10 9 the inductance is expressed in henrys. Problem 69. Prove that there is no flux outside of the sheath, the flux set up there by the current in the sheath being exactly neutralized by the flux set up in the same space by the oppositely directed current in the inner conductor. Inductance of a Transmisson Line. Let a transmission line be represented as in Fig. 128 by 2 conductors, A and B, of radius r. Let the distance between their centers be D. Each conductor surrounds itself with flux lines, the directions of which are indicated by arrows. The flux through any zone of width, dx, between the conductors, due to the current in A, is SIMPLE PROBLEMS IN ELECTRO-STATICS 161 where x is the distance of the zone from the center of A, and F x is the m.m.f. due to A. Similarly, the flux through dx, due to the current in B is The flux due to both A and B is then / / The inductance due to the / I / interlinkages of the conductors f I j with the flux between them is \ 1 \ then, since F x = I in this case, \\ V , D-r 4 M . D - r , = 4/i log ^ ' cm. or j^ log - henries per cm. To determine the total inductance per centimeter length of circuit, that due to mterlinkage within the material of each con- ductor must be added. This has been found (74) to be ^ for 2 each conductor. Therefore, the total inductance is L (total) = 4;u log -- + n cm. per cm. of circuit (75) In practical formulae, this becomes L = 0.000015 + 0.00014 logio^- in henrys per 1000 ft. of wire, not 1000 ft. of circuit. Note that if the capacity between transmission lines is given in farads and the inductance in henrys ~7ffi ^ on ^ verv less than the velocity of light which is 3 X 10 10 cm. per sec. or 187,000 miles per second. Problem 70. Explain the effect of increasing the size of the wire on the inductance of a transmission line. Similarly, explain the effect of increasing the distance between the wires. 11 CHAPTER XXIV DISTRIBUTED INDUCTANCE AND CAPACITY In the electric and magnetic problems dealt with so far it has been assumed that the electro-static and magnetic fields propa- gate with infinite velocity. In other words, it has been assumed that the instantaneous values of the currents and e.m.fs. are the same at all points of the circuit. This of course is practically true except in very long transmission lines, since the propagation of the electric and magnetic fields in a dielectric such as air is the same as that of light, or very nearly 3 X 10 10 cm. per sec. or 187,000 miles per sec., and along a transmission line it is re- tarded only a small percentage due to the fact that the current is not confined to the surface of the conductor. Assuming, however, that the transmission line is very long, say 300 miles, then the time interval between, say, the maximum , , value of the current at the beginning dl : "1 r~^* an d the end of the line is evidently ;Hj20 sec -> corresponding in a 60-cycle - system to approximately one-tenth of one cycle, or, approximately, 36 in FIG ' 129 ' time phase. It is thus seen that in a long transmission line not only do the instantaneous values of the currents and e.m.fs. vary from instant to instant, but at a given instant the values of the currents and e.m.fs. are different at different points of the line. This problem has been treated very completely by many authorities. The simplest solution appears to be that by STEIN- METZ, 1 which is largely followed in the succeeding paragraphs. Let Fig. 129 represent a long transmission line. Let r = re- sistance per unit length of line, X Q = reactance per unit length of line, Power factor at the generator = ^ry Power per phase of the load = P. Volt-amp, per phase of the load = el. Power factor of the load = -* p -* P Efficiency of transmission Voltage regulation = (Eo e) -f- e. Plot the voltage and current vectors for both ends of the line. Solution. Resistance of No. 000 B. & S. wire, from tables, = 0.0605o>/1000 ft. at 60F. .'. r = 0.0605 X 5.28 X 200 = 64 ohms. Inductance = 0.00014 logic ^-^ = 0.00014 lo glo 304 ' 5 Q 5 ~ ' 52 per 1000 ft., where r = radius = 0.5202 cm. .'. L/1000' = 0.00014 logio 585 = 0.00014 X 2.767 = 0.0003874. L = 0.0003874 X 5.28 X 200 = 0.409 henry. X = 0.409 X 377 = 154.2 ohms. C/1000' = ' r = 0.0074 X 0.361 = 0.002672 micro-farad logio jr- C = 0.002672 X 5.28 X 200 = 2.82 micro-farads. 6 = 2T/C = 377 X 2.82 X 10~ 6 = 0.00107 W corona loss per wire 200,000 nnnn oo, 9 ~ * ~ (voltage to neutral) = (72,250) * " 1 The voltage to neutral on a balanced three-phase system is the line voltage divided by \/3 = 7-=^- l.f O The current supplied from the generator is found from (88) to be: (V7\ I V 7\ 1 + "27 + Y = (10 ~ J ' 50) I 1 + T) + e Y- YZ = gr - bx +j(gx + 6r); Y = g -f- j6; e = 72,250. /. -^ = - 0.081 +;0.037. YZ 1 +-2~ - 0.919 + J0.037; eY = 2.741 + J77.3. Substituting values, I = (100 -j50) (0.919 +J0.037) +2.741 + J77.3 = 93.75 - J42.2 -f- 2.741 + j'77.3 = 96.49 + J35.1 = 102.5 amp. DISTRIBUTED INDUCTANCE AND CAPACITY 167 The voltage at the generator terminals is obtained in a similar way, and is o = e(l + ^r) + IZ = 72,250(0.919 +;0.037) + (100 - J50) (64 = 80,625 + .;14,928 = 81,200 volts. The power per phase at the generator is, by "telescoping" E I , Po = e i + e V = 80,625 X 96.49 + 14,928 X 35.1 = 8200 kw. The apparent power input to the line is #0/0 = 81,200 X 102.5 = 8325 k.v.a. at the generator. The power factor at the generator is Po 8200 P.F.o = 8325 0.985. The power supplied to the load is p = e i = 72,250 X 100 = 7225 kw. Load FIG. 131. The apparent power supplied to the load is el = 72,250 X VlOO 2 + 50 2 = 72,250 X 111.9 = 8060 k.v.a. The power factor of the load is .,' ; r.*--ig-". P 7225 Efficiency of transmission = p- = OOQA = 0.882. 168 ELECTRICAL ENGINEERING 81,200 - 72,250 Regulation = - 72 25Q = 12.4 per cent. The vectors E , 7 , e and 7 are plotted to scale in Fig. 130. Problem 72. Consider a circuit as shown in Fig. 131. Let the con- stants be: r = 0.01 x = 0.02 ri = 0.01 xi = 0.02 r = 0.01 x = 0.002 When the load voltage is e = 1, and the load current is 7 = 1+ Q.5j, find the generator voltage, current, power factor, and the voltage and current of the branch (r , So). CHAPTER XXV NOTES ON THE MATHEMATICS OF COMPLEX QUANTITIES This chapter is inserted in order that the common mathe- matical operations shall be kept fresh in mind by review and frequent practice. It is very desirable that the student shall possess and retain facility in common, though not always fre- quent, operations. For instance: Solve Vo.008, using log tables. Solve 6-- 216 , using log tables. Differentiate y = ax n ; y = ae~ ax ; y = sin x; y = cos x; u y u uv; y = - Find the log, and differential, of 4 3j. Find \/-3]. Representation of Complex Quantities. The general expres- sion for a complex quantity is A = ai -f j2. The numerical en FIG. 132. FIG. 133. value, or modulus, of the complex is A = \/ai 2 + a 2 2 and the vectorial angle is tan" 1 a = . These various quantities may be represented as in Fig. 132. Then a : = A cos a; a 2 = A sin a, whence A = A (cos a + j sin Q) = Ae 3 ' a , the latter relation being proved later. Addition of Two Complex Quantities. Let C = A + B (Fig. 133). 169 170 ELECTRICAL ENGINEERING Then C = ai + ja 2 + 61 + j& 2 = ai + bi -f j(2 + W> and V (ai + 6]) 2 + (a s + 6 2 ) 2 . Multiplication of Two Complex Quantities. Let C = A XB. C = (ai+ja 2 )(&i+j& 2 )' and C = Division of Two Complex Quantities. A Let C = - 4- j2 __ a\l>\ and Tan 7 = Similar processes may be carried out when the complex quantities are expressed in polar coordinates. Multiplication. C = AB = a(cos a + j sin a)6(cos + j sin 0) = A(cos cy cos + j sin a cos + cos aj sin |S sin a sin ft) = AB(cos (a + 0) + j sin(^ + ). Involution and Evolution. A 2 = A V' a = A 2 (cos 2 a + j sin 2a). A n = A n (cos na + j sin na). A^ = A^ (cos - + j sin -) (89) \ n nl MATHEMATICS OF COMPLEX QUANTITIES 171 Since cos a = cos (a + 2?rp) and sin a = sin (a + 2irp) where p is any integer, the simple complex expression should be written : A = A [cos (a + 27rp) + j sin (a + 2irp)], where there is any question about the number of different solutions. In evaluating such expressions, a is in radians. Sin X and cos X may also be written as series, 1 in which cr Sm # = r 3 * . -f- FT Cos x = 1 - ry + r^ - (90) Example. Calculate, from series expression, the value of sin 2. Since the angle must be expressed in radians, 2X27T 7T * = "360T = 90 radians ' Substituting this value into the series, o _ sm J " 90 6 X 90 3 = 0.0349. 120 X 90 5 The Roots of a Complex Quantity. Using the more general expression, Eq. (89) may be written: 2irp cos + j sin (91) n n 1, 2, 3, 4, etc., and solve, continuing To find the roots, put p until repetition begins. Example. Find \/l = where A = 1 = 1 + jo. A = A (cos a + j sin a) = 1 ; a = l',n = 4; tan a = y = Tabulating, and supplying values to (91) p 01234 2wp 4 cos -1 -i sm jO jl JO jl JO v 7 A 1 j -1 -j 1 The roots are represented as vectors in Fig. 134. 1 Developed by MACLAURIN'S theorem. -3 FIG. 134. 172 ELECTRICAL ENGINEERING Exponential Representation of Complex Quantities. The ex- ponent e" may be written as a series known as the exponential series, developed from MACLAURIN'S theorem. Thus, u u 2 u 3 Let u = jB. Then, * _ , , # , !! + 1 + I 2 + ' ' ' je P _ JP ^ -I- T~ I 1 '" O I O I 'I A "(i-f+S) < These two component series are seen to be those of the sine and cosine (90). Hence (92) may be written: j0 = cos 6 +j sin e (93) Since A = A (cos a + j sin a), substituting from (93), A = Ae ja . Thus, a third form of writing the complex quantity, A, has been developed. This last may be extended by letting A = e ao . Thus, A o Va _. ao+jat jtl e c C in which the exponent is complex. Differentiation of a Complex Number or Vector. Let A = Ae ja . then dA = Aje ia da + J a dA ' = e ja [Ajda + dA] (94) Logarithm of a Complex Number or Vector. Cdu Iqgn-J We have log A 4= f - J 4' MATHEMATICS OF COMPLEX QUANTITIES 173 from (94), rt>"Ajda . CdA^ r rdA 1 * A = J ~AfT + J "At* == J ^ + J T = j(a + 27rp) + log A. The logarithm of a vector has thus an infinite number of values. / / CHAPTER XXVI THE TRANSFORMER The alternating-current transformer is used to change electric energy from one voltage to another. This is done by interlinking two electric circuits having different numbers of turns with the same magnetic alternating flux. If the two circuits enclose exactly the same flux it is evident that the voltages induced in the windings will be proportional to the numbers of turns. If, however, as is the case, the flux is not exactly the same for each circuit, the ratio is slightly affected and, as will be shown later, the secondary voltage has a value differing slightly from what the ratio of turns would demand. When one circuit is connected to an alternating e.m.f., the other circuit being open, a current flows in that circuit (Fig. 135). This current is called the no-load or the exciting current, and may be assumed to consist of two components, one of which Inf FIG. 135. FIG. 136. FIG. 137. supplies magnetism to the core and is called the wattless com- ponent, while the other supplies power for hysteresis and eddy current losses and is called the power component. These component currents of the exciting current may be rep- resented as flowing in a circuit of resistance and inductance in parallel as in Fig. 136, where e is the e.m.f. which sets up these currents. They may be represented vectorially, as in Fig. 137. In the latter representation i m , in quadrature with e, produces the flux 0, but no power; 4, in phase with e, supplies the core loss. The exciting current, 7 o, lags behind e by an angle tan -1 -r-, ih It is not strictly correct to represent the core loss by a resist- ance r, Fig. 136, with varying e, for part of the core loss is pro- portional to e 1 - 6 and part to e 2 . 174 THE TRANSFORMER 175 Neither is it correct to assume that the magnetizing component is proportional to the e.m.f., since the magnetization curve is not a straight line. However, in most cases, the variation of e is slight, and proportionality may be assumed without appreciable error. The Transformer Diagram. The relations of voltage, current and flux which exist in a transformer under normal operation are shown with great clearness by the aid of the transformer diagram. represents the flux that interlinks with the primary and second- ary of the transformer; e t - is the e.m.f. induced in the primary and secondary windings (assuming the same number of turns in each). This e.m.f. is 90 in time behind the flux, as is seen from Fig. 139 and by the following simple proof: If $ = m sin ( then N d N FIG. 139. 1 2 is the secondary or load current which in this particular diagram is shown lagging behind the induced e.m.f. 7 2 r 2 and 7 2^2 are respectively the e.m.fs. consumed by the secondary resistance and reactance, /2r 2 being in phase with 7 2 and 7 2 z 2 being 90 ahead of 7 2 . 7 2 z 2 is the e.m.f. consumed by the secondary impedance, which subtracted vectorially from e t gives E 2 as the secondary terminal voltage. The primary current may be assumed to consist of three com- ponent parts: the first I\, which corresponds to the secondary current and is equal and opposite thereto; the second 7 m , which is 176 ELECTRICAL ENGINEERING the magnetizing current producing the flux and is in phase with the flux; and the third Ih, which is the power loss current due to the core loss and is in quadrature to the magnetizing com- ponent, that is, in phase but opposite to the induced e.m.f. e t . I m and I h combine in 7 00 which is the exciting current. To overcome the induced e.m.f. e t - in the primary winding an impressed e.m.f. e { is required. To overcome the resistance and reactance drop in the primary windings an e.m.f. I\z\ needs to be supplied. Thus the pri- mary impressed e.m.f. E\ is the vector sum of these. 0i is evidently the angle between the primary current and e.m.f. 6 2 is the angle between the secondary current and e.m.f. The total primary current, /i = I'\ -f- 7 o. In phase with /i is the voltage, I&1, consumed by the primary resistance, r j; and at right angles ahead of /i is the voltage, I\Xi t consumed by the self-inductance of the primary coil. These two voltages combine to form /iz 1; the voltage consumed in the primary of the transformer. The total impressed primary voltage, EI, is the sum of IiZi and d. The angle 0i is the phase angle between EI and /i. The transformer diagram is obviously not suitable for accurate calculation. For this purpose, another de- gj . velopment will be made. .4, j lf l Let there be two mutually inductive coils, one of them, called the primary, having NI L N turns, TI ohms resistance, and LI henrys in- ductance, while the similar quantities of the FIG. 140. other, or secondary coil, are N* y r 2 and L 2 respectively. Then, in Fig. 140, if the secondary current 7 2 = 0, the primary impressed voltage, ei = i l r l -f LI -p where e\ and i\ are instan- taneous values of voltage and current, and Li is assumed con- stant. If a secondary current flows, there will be induced in the secondary an e.m.f. e t = - L 2 -J 2 - The secondary induced e.m.f. perturn=-f -*?. Nt N 2 dt If it be assumed that there is no leakage, that is, that all the magnetic flux links with both the primary and the secondary THE TRANSFORMER 177 coils, then the induced e.m.f. in the primary due to 7 2 must be N l T di* ~N~ 2 L ^t' Then, The sign of the last term changes from to + because the in- duced e.m.f. must be overcome, or balanced, by an e.m.f. of the opposite sign. But JV2 L 2 = \LA for, from fundamental relations, ^1 = T7^r> and #i = Substituting, L! = Similarly, L 2 = Then the ratio Li _ L 2 ~ whence (96) then becomes *'r \/rr F 2 ^ 2 Nf ^ i ^ 2> & 61 = i\r\ -f~ LI -37 -f- \LiL 2 j. at at Let \/L]L 2 be denoted by 3f . Then, Similarly for secondary, _i_ / ^ 2 4. M l = n (98) since no e.m.f. is impressed on the secondary coil. The constant, M, is called the coefficient of mutual induction, and may be defined as the number of interlinkages of flux with both coils of a mutually inductive circuit when unit current is flowing in one of the coils. 12 178 ELECTRICAL ENGINEERING Mutual inductance, like self-inductance, is measured in henrys. It does not always follow that M is equal to \/LiL 2 . In fact, that condition is attained only when no magnetic leakage exists, which never occurs. If part of the flux set up by the primary does not interlink with the secondary, that part constitutes the primary leakage flux. Similarly, when current flows in the secondary, some secondary leakage flux is set up. Whenever there is leakage flux, M Eqs. (97) and (98) hold at all times provided the proper value of M is supplied, and M is usually about 95 per cent, of v LiL 2 . Equivalent Transformer Circuit. The differential equations given above are not readily used, but, fortunately, STEINMETZ has evolved a simple treatment involving a diagram of simple series and multiple circuits, which, while not showing the physics of the phenomenon, lends itself to very simple and quite accurate treatment. He represents the transformer by a circuit which is shown in Fig. 141. r l % r a J a t ^ t ri. f *oo| Si l&oo M 1 I 1 i V FIG. 141. Let the secondary or load current be 7 2 = iz + ji't, and let the secondary terminal voltage be e 2 , the zero vector. Then the secondary induced e.m.f . E t = e* + hZ 2 = e z + itfz The exciting current is /oo = EiY Q = (i -f je'i) (goo + j&oo) The primary current is /i = /2 + /oo = ^2 + IO The impressed voltage is THE TRANSFORMER 179 From these values may be obtained: power output = e^iz, power input = &iii -f e\i'\ t f . . , power input e\i\ + d\i\ power factor at primary terminals = rr- = ' T > volt-amp. EJi regulation = 2 62*2 efficiency In using these equations r*i, r 2 , Xi and x 2 are positive, & o is negative because the magnetizing circuit is necessarily inductive, iz is negative for lagging, positive for leading current. Transformers are rated on the basis of kilovolt-amperes, not kilowatts. Example of Transformer Calculation. Given a 2200 to 220- volt, 60-cycle, 50-kv.a. transformer, in which r*i = 0.97, r 2 = 0.0097. Assume that on test 98.5 volts on the primary produces full-load current in the short- circuited secondary (142, a). At no-load, with the normal voltage (220) impressed on the secondary, the primary circuit being open, the watts input are WQ = 1000, and the exciting current / O o = 12.25 amp. (Fig. 142, 6). The percentage rl drop in the primary is 22.7 (6) FIG. 142. 22.7 X 0.97 -2200" = 0.01 == 1 per cent. where 22.7 is the normal primary current. In the secondary, per cent, rl drop = = .01 = 1 per cent. The total impedance, calculated from the short-circuit test (142, a), is QO K Z Mal = 2277 4.35 ohms. 180 ELECTRICAL ENGINEERING Total per cent, impedance drop, referred to the primary voltage, is = - 0448 = 4 ' 48 P er cent ' 10 IOOOO5 IOIOO'-HOI>OOOOOOO5 O Ot^O OOOOOOOOO rH OOOOOO 1 1 lOtNt^CllN lOrH CO CD (M O T^ i t CO O N. l>t^Oi CO 1 ^ COrHOCOO^O T^ T^t^i-tC> I o 0> w CHAPTER XXVII HYSTERESIS AND EDDY CURRENT LOSSES Hysteresis Loss. The hysteresis loop is interesting in that it indicates by its area directly the work done on the electromagnet per cycle of change of current. The work done in an electric circuit has been shown to be feidt. If T is the time of the cyclic variation of current then, C T W = I eidt, is the work performed during the cycle. J Q But the induced e.m.f. in a winding of N turns is e = jgg -=r> where -^ is the rate of change of flux. .'. W = I TT^ -j7 dt. But (j) = SB, where S is the cross- sectional area of the magnetic circuit in square centimeters. . W . . rx&Ma Jo 10 8 dt C Also the magnetizing force is: QAiriN tf = -_ , where i is given in amperes. Thus, A7 IH iN = 0.47T' and, substituting this value, r SIH as si r J 0.47T xW* ~di c ' WxT*h HdB ' But SI is the volume, 7, of the magnetic structure. Thus, V C T W = - HdB. 10 7 X 47rJ But HdB is the area of the hysteresis loop corresponding to maximum density, B, as seen from the loop. The work is given in joules. 186 HYSTERESIS AND EDDY CURRENT LOSSES 187 STEINMETZ found that the hysteresis loss in watts could be expressed (approximately) by the following equation: W 10 7 where V is the volume and rj is a constant which depends upon the quality of the iron. The equation shows that the loss is proportional to the 1.6 power of the maximum density and directly proportional to the frequency. In centimeter measure- ments ri varies from 0.001 to 0.002 in ordinary sheet iron and may be 10 times as great in tempered steel. In the best silicon steel it is 0.0006, which corresponds to 0.54 watt per Ib. at 60 cycles and a density of 64,500 lines per sq. in. or 10,000 lines per sq. cm. Eddy Current Loss. Eddy currents differ in no way from other currents, and the loss of power by them is therefore i 2 R or if E is the e.m.f. causing the current and Z is the impedance of the path, then, and the loss is It follows, then, that the loss is proportional to the square of the e.m.f. or, what is equivalent, to the square of the maximum density and to the square of the frequency, since the e.m.f. itself is proportional to the frequency of flux variation and the maximum density. Even in the simplest cases it is difficult to calculate the loss since the distribution of the flux and, therefore, the e.m.f. in different parts of the material is often very complex. Consider as an illustration the simple case of eddy current loss in transformer steel. The cores are built up of laminations in such a way that the flux path is divided up into a number of elements each having the section of the edge of a lamination and following parallel, or as nearly so as possible, to the sides of "^j^ 145 the laminations. With the flux entering, as is shown in Fig. 145, currents will flow as indicated by the dotted lines. The current flowing 188 ELECTRICAL ENGINEERING through a section of area l\dx encloses a flux which is r 0, where u is the flux passing through the entire area of one lamination (assuming uniform flux density). 1 The effective value of the e.m.f. induced is 4.44 X flux X turns X frequency _ \/27r2o;flidx where p is the specific resistance of the material. .'. t' 2 r in the elementary circuit is x 2/p 4 p 2 h(dx) and the total loss is p ' 4***f*l0*(dx) Jo IVWp 6 X 10 16 Z P Since the volume is Hid, the loss per cm. 3 is W TT^hd 1 TT 22 X F 6 X 10 16 Z P Z X W ~ 6 X 10 16 / 2 p But = 5 X W. /. 2 = and W_ Tr 2 BH 2 d 2 f* V ~ 6 X 10 16 Z 2 p ~ 6 X 10"p Watts * p for sheet iron is about ^ ohms. 1 For a more complete discussion see " Advanced Electrical Engineer- ing." CHAPTER XXVIII WAVE DISTORTION IN TRANSFORMERS If on a transformer containing no iron a sine wave of e.m.f. were impressed at its terminals, the flux and the exciting current would also follow sine waves. With the introduction of iron, however, while the flux values would still follow a sine wave, or very nearly so being distorted only due to the ohmic drop of the distorted current the exciting current wave would necessarily be considerably distorted. Its shape is shown in Fig. 148, which is derived from the hyste- resis loop given in Fig. 147. Conversely, if by some arrangement the exciting current were made to follow substantially a sine wave, the flux wave, and therefore the wave of voltage across the transformer, would be greatly distorted. This distortion in current or e.m.f. waves is of considerable importance in connection with the grouping of transformers in a three-phase system, as will be seen later. At present, however, only the condition in a single-phase transformer will be studied. A representative hysteresis loop is shown in Fig. 147, which was obtained from actual tests with a sine wave of impressed e.m.f. The test data are recorded in Table VI. If the effect of the ohmic drop be neglected, then the impressed and counter, or induced, e.m.f. are the same numerically and where N is the number of turns and < is the flux. With a sine wave of flux = $> m sin co, dt -^ = 4> TO co cos at. cos a)t = E m cos The induced e.m.f. has its negative maximum when the flux begins to rise, and lags behind the flux by 90 time degrees. Thus 189 190 ELECTRICAL ENGINEERING the impressed e.m.f ., E, which is equal and opposite to the induced e.m.f., leads the flux by 90 (neglecting the ir drop), Fig. 146. If instead of being a sine wave the flux were distorted and yet symmetrical, it would be represented by FOURIER'S series of odd harmonics, thus: = 3> lm sin at + 3wi sin (3o>Z + a) /. e . = N -ir = &i m w cos ut The e.m.f. wave would be relatively more distorted than the flux wave as is evident from the coefficients of the different trigonometric terms. sn cos -fa)... FIG. 147. When a hysteresis loop is given, if either the flux wave or ex- citing current wave is known, the other may be at once obtained. For example, let the flux wave be assumed to be sinusoidal. TABLE VI. HYSTERESIS LOOP DATA Ord. Abs. Aba. 0.0 0.5 -0.5 0.2 0.56 -0.43 0.4 0.63 -0.32 0.6 0.71 -0.18 0.8 0.82 0.08 0.9 0.9 0.35 1.0 1.0 1.0 EXCITING CURRENT DATA Time Flux ioo 0.0 0.5 10 0.174 0.55 20 0.34 0.6 WAVE DISTORTION IN TRANSFORMERS 191 The exciting current data are obtained from the hysteresis loop by reading off the current values corresponding to the flux values which have been taken at uniform intervals along the flux wave. Thus, at on the flux wave $ = 0. This value of <, on the hysteresis loop, corresponds to i 00 = 0.5 amp. At 10 on the flux wave, < = 0.174. This value on the loop corresponds to z'oo = 0.55, etc. Data for the exciting current are given in Table G/T. It should be noted that the flux naximum and current maximum always >ccur at the same instant. / *- The phase relations and character- istic current wave shape for a sine wave of flux are shown in Fig. 148 The im F IG pressed voltage wave leads the flux by 90. The scales to which the waves are plotted are quite in- dependent of each other, and should be so chosen as to exhibit the waves most clearly. When the induced e.m.f. is not a sine wave, the flux wave is also distorted. In this case the impressed e.m.f. -N**. ~ * dt Transposing, edt where N is the number of turns. Hence ft - <2 />2 I &- A^d0. ./ = i ^i If ti is chosen as the time when is zero, and tz is the time when is maximum, then AC-* ^ N N !*- ioi^ = ioi Ji = (i t/0 This equation shows that the maximum value of the magnetic flux or flux density in which the electrical engineer is very much interested, since it determines the magnetizing current and core loss is proportional to a certain area of the e.m.f. wave, and it remains to determine where this area is located. When the flux is a maximum then - is zero; thus e is zero. 192 ELECTRICAL ENGINEERING The value of tz is therefore easily ascertained, as is shown in Fig. 149. The ordinate through ti must bisect the e.m.f. wave in order that the flux wave be symmetrical, as can also be seen by slight consideration, since the flux wave must be symmetrical above and below the zero line. Thus, in finding the flux wave, the first step is to bisect the area of the e.m.f. half-wave, which gives the posi- FIG. 149. tion of ti and the zero of the flux wave. Problem 76. From the following readings on a distorted e.m.f. wave obtain and plot the flux and current waves. NOTE. Choose a scale to give 6 Hysteresis loss in wave B ~ \av. e.m.f. of B/ 1 This is demonstrated on p. 228, Chap. XXXII. 8 See Chap. XXIX. WAVE DISTORTION IN TRANSFORMERS 195 By definition, Form factor (f .f.) = effective e.m.f. average e.m.f. . Hysteresis loss in A _ rf.f. (B)"| L6 ' 'Hysteresis loss in B Lf.f. (A)J Therefore, the higher the form factor the less the core loss. The form factor of a sine wave is 1.1. In general that of a flat- top wave is less; of a peaked wave, more. Wave A (Fig. 152) has maximum core loss. Wave B has minimum core loss. FIG. 152. CHAPTER XXIX DISTORTED WAVES It is often necessary to express a distorted wave in the form of an equation. This can readily be done since it has been found that any periodic univalent curve can be expressed by a series of terms involving a constant and sine and cosine terms. That is, y = a + ai cos + 2 cos 26 + + a n cos nB + 61 sin + 6 2 sin 20 + + 6 n sinr*0 (103) represents any distorted wave in which for every value of abscissa only one ordinate exists, provided that the abscissa is so chosen that the curve repeats itself at a value of = 2?r, i.e., the wave is periodic. Obviously, if the distorted wave is given graphically it is always possible to read off the ordinate corresponding to each abscissa (Fig. 153). 2 7T FIG. 153. The problem then resolves itself into finding the coefficients o, i, n, &o, &i, &in (103). To do this a mathematical transformation has been worked out involving convenient integrations and the fact that sines and cosines have the same values at = as at = 2?r or any multiple of 27r, that is, 2?rn, where n is an integer number. To find a integrate Eq. (103) between and 2?r. Thus, yds r^ I - I ade + rzr I cos BdB + cos nOde -f I 61 sin 6d0 + 196 1 ri* I a n - b n sin n0d0. DISTORTED WAVES 197 From what has been said above, all integrals except the first must be zero. Thus j ydO = I a dB = a (2ir - 0) = 27ra . 1 F* .'. a = -^~ yds. ^ But I yds is the area of the curve during one complete period and 2ir is the abscissa. .'. a is the average value of all the ordinates, or the average value of y. To determine any other coefficient, for instance a, Eq. (103) is multiplied by cos nB and integration is again carried out be- tween limits and 2ir. In this case it is also remembered that the integral over one period of any product of sine and cosine terms is zero. rr r2* r2* y cos nBdB = a I cos nBdB + i J cos nB cos BdB + a n I cos 2 nBdB + fei I cos nB sin BdB- + fe n I cos nB sin n&dB. All these integrals on the right-hand side must be zero with the exception of cos 2 nBdB, and this integral, as is readily seen, is = TT. I C 2v .'. a n = - I y cos nBdB. But J*y cos nBdB is the area, not of the original curve, but of another curve which is obtained by multiplying each value of y by the particular value, at phase angle B } of cos nB. Since that area is divided by TT the integral must be just twice the average of the instantaneous values of ?/, multiplied by cos nB. .'. a n = 2X avg. of y cos nB between and 2ir. In a similar way all values of fe are obtained so that, 6 n = 2X avg. of y sin nB from to 2ir. 2 .'. a = avg. (y) 198 ELECTRICAL ENGINEERING ai = 2X avg. (y cos 0) a 2 = 2X avg. COS 20)0' a 3 = 2X avg. (y COS 30)1' a n = 2X avg. (y COS I 2 ' n0)!o 2 61 = 2X avg. (y sin ?) 62 == 2X avg. (y sin 20 )o b n = 2X avg. (y sin n0)l It should be noted that dividing the curve up, say every 10 from to 360, 37 readings are obtained. It is better then to use 36 and to take the average value of the values at and 360 instead of using both of them. In a symmetrical wave only those harmonics can exist, which, with an increase of the angle by 180 or TT, reverse the sign of the function. This is only the case when n is an odd number. Since, if n is 2, 4, 6, etc., then increasing the angle by TT means 27r, 4?r, 6V, etc., and the values of the sine and cosine are the same for a, (a + 2ir), (a -f 4?r), etc., whereas if n = 1, 3, 5, etc., we get, TT, 3r, STT, in which the sign of the function reverses. If sin a is positive, then sin (a + TT) is negative. If cos a is positive, cos (a + v) is negative, etc. Thus, for symmetrical waves such as are given by alternators under stable conditions, the trigonometric series becomes : y = ai cos + a 3 cos 30 + a 5 cos 50 +. . . -f- 61 sin + 6 3 sin 30 + &5 sin 50 + .... Obviously, in that case, it suffices to analyze one-half a wave only. 1 Problem 78. Plot the wave, e = Ei sin 6 + E 3 sin (36 + a), for E l = 1 E z = 0.5 a = 30, and analyze the wave, proving that the analysis gives the original equation. Show also that no 5th harmonic exists. 1 For a more complete discussion of this method of wave analysis see STEINMETZ'S " Engineering Mathematics." DISTORTED WAVES 199 Tabulating : 6 10 20 30 40 50 60 70 80 90 Ei sin 0.0 0.174 342 50 643 766 866 94 935 1 38 + a Sin (30 + a) tfasin (39 + a)... 30.0 0.5 0.25 25 60.0 0.866 0.433 607 90.0 1.0 0.5 842 120.0 0.866 0.433 933 150.0 0.5 0.25 893 180.0 0.0 0.0 766 210.0 -0.5 -0.25 616 240.0 -0.866 -0.433 507 270io -1.0 -0.5 485 300.0 -0.866 -0.433 567 100.0 110.0 120.0 130.0 140.0 150.0 160.0 170.0 180.0 Ei sin 0.985 0.94 0.866 0.766 0.643 0.50 0.342 0.174 0.0 36 + a 330.0 360.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 Sin (30 + a) -0.5 0.0 0.5 0.866 1.0 0.866 0.5 0.0 -0.5 Et sin (36 + a).. . -0.25 0.0 0.25 0.433 0.5 0.433 0.25 0.0 -0.25 e 0.735 0.94 1.116 1.2 1.143 0.933 0.592 174 -0 25 Analysis. a must be zero because the wave is symmetrical above and below the center line. The coefficients of the fundamental cosine and sine waves are found from ai = 2 X avg. e cos 0, bi = 2 X avg. e sin 6. e Cos e e e cos Sin $ e sin 1.0 0.25 0.25 0.0 0.0 10 0.985 0.607 0.598 0.174 0.1057 20 0.94 0.842 0.792 0.342 0.288 30 0.866 0.933 0.808 0.5 0.466 40 0.766 0.893 0.685 0.643 0.575 50 0.643 0.766 0.493 0.766 0.587 60 0.5 0.616 0.308 0.866 0.534 70 0.342 0.507 0.173 0.94 0.477 80 0.174 0.485 . 0844 0.985 0.478 90 0.0 0.5fc7 0.0 1.0 0.567 100 -0.174 0.735 -0.128 0.985 0.725 110 -0.342 0.94 -0.322 0.94 0.885 120 -0.5 1.116 -0.558 0.866 0.966 130 -0.643 1.2 -0.772 0.766 0.920 140 -0.766 1.143 -0.875 0.643 0.735 150 -0.866 0.933 -0.808 0.5 0.466 160 -0.94 0.592 -0.556 0.342 0.202 170 -0.985 0.174 -0.171 0.174 0.0303 180 -1.0 -0.25 -0.25 0.0 0.0 200 ELECTRICAL ENGINEERING The sum of the 18 cosine readings, using the average of and 180 as one, is - 0.2486 and the average value is - 0.0138. Thus, ai = 2 X avg. = - 0.0276. Similarly, the sum of the sine readings is: 9.007 The average is 0.5004, Thus, 61 = 2 X avg. = 1.0008. The coefficients of the 3d harmonics are found from, a 3 = 2 X avg. e cos 30, 6 3 = 2 X avg. e sin 36. e 30 Cos 30 e cos 30 Sin 30 e sin 30 1.0 0.250 0.0 0.0 10 30 0.866 0.525 0.5 0.304 20 60 0.5 0.421 0.866 0.730 30 90 0.0 0.0 1.0 0.933 40 120 -0.5 -0.447 0.866 0.774 50 150 -0.866 -0.664 0.5 0.383 60 180 -1.0 -0.616 0.0 0.0 70 210 -0.866 -0.439 -0.5 -0.254 80 240 -0.5 -0.242 -0.866 -0.420 90 270 0.0 0.0 -0.1 -0.567 100 300 0.5 0.368 -0.866 -0.637 110 330 0.866 0.815 -0.5 -0.470 120 360 1.0 1.116 0.0 0.0 130 390 0.866 1.040 0.5 0.600 140 420 0.5 0.571 0.866 0.990 150 450 0.0 0.0 1.0 0.933 160 480 -0.5 -0.296 0.866 0.513 170 510 -0.866 -0.151 0.5 0.087 180 540 -1.0 0.250 0.0 0.0 The sum of the 18 cosine readings is 2.251. The average is 0.125. .'. a 3 = 2 X avg. = 0.25. The sum of the sine readings is 3.899. The average is 0.2165; .*. 6 3 = 0.433. The exercise of proving that no 5th harmonic exists is left for the student. Summing up the values already obtained, the equation may be written : e = - 0.0276 cos 6 + 0.25 cos 30 + 1.0008 sin - 0.433 sin 30 which is, approximately, y = sin + 0.433 sin 30 + 0.25 cos 30. DISTORTED WAVES 201 The second and third terms may be combined or added, being in quad- rature, by the vectorial method in which where Thus, A sin 6 + B cos = \/A 2 + B 2 sin (0 + a), tan a = A A. 0.433 sin 30 + 0.25 cos 30 = Vo.188 + 0.0625 sin (30 + a) = 0.5 sin (30 + a), where a = tan" 1 Q-TOQ = 30. The complete wave is, therefore, e = sin -{- 0.5 sin (30 + 30). The wave is shown plotted in Fig. 154, in which also the component waves are indicated by the dotted lines. FIG. 154. CHAPTER XXX MECHANICAL STRESSES IN TRANSFORMERS It is recollected that a mechanical force is exerted on a con- ductor carrying current if it is placed properly in a magnetic field, the force being 1 dyne per cm. of conductor per abamp. in a field intensity of 1 line per sq. cm. provided the field is at right angles to the conductor. Referring to Fig. 155, which represents the cross-section of a transformer, it is evident that the main flux which interlinks with both" the primary and the secondary windings and is confined to the iron does not cut through any part of the windings carrying current, but that the leakage flux more or less completely cuts the windings and therefore is responsible for a force which tends to warp the coils out of shape and thus to damage them. The deter- mination of the mechanical stresses resolves itself therefore largely into the calculation of the leakage core flux or leakage inductance of the transformer. To calculate the leakage induc- tance of the secondary coil, con- sider this made up of the interlink- . ages of flux with turns in the space FIG. 155. occupied by the secondary coil itself, plus the interlinkages of the flux between the coils with all of the secondary turns. Similarly with the primary. Approximation of the Leakage Inductance of the Secondary. - turns, where In Fig. 155, a portion of the coil of depth x, has a is the total depth of the coil, and N 2 is the total number of secondary turns on 1 leg of the transformer. The magnetomotive force of this part of the coil is t r *r x m.m.f x = I 2 N 2 a The flux which this m.m.f. produces is flux = 202 MECHANICAL STRESSES IN TRANSFORMERS 203 where p is the reluctance, and p = = - when we con- area max sider only the flux which passes through the small area of width dx and length m. m is the length of a turn at distance x in Fig. 155. It is almost impossible to determine accurately the length Z . It is the equivalent length of the lines of force which going through section mdx return upon themselves. Part of these lines can be readily traced. They go almost straight across the trans- former windings of length h; then they spread apart, and the equivalent length, as a result, is relatively short. Then, the majority of the lines enter the iron and their reluctance is insig- nificant. Some, however, enclose the winding that is outside of the iron and these meet with considerable reluctance. Therefore, it might be fairly conservative to assume Z , the equivalent length, as I the height of the "window" of the transformer. If mz is the mean length of a secondary turn, this may be sub- stituted for m, thus I p ' Then the flux in any elementary band, dx, is m 2 dx x d' This flux interlinks with N z turns. Therefore, the interlink- /> /y ages with the flux = ^irl^Nz "**""" - N%, and the inductance due to the interlinkages within the space occupied by the coil is (104) 4 To determine the inductance due to the flux in the gap between coils, consider Fig. 156 which shows a section through one side of the coils. The current is oppositely directed in the two coils, as indicated by dots and crosses. On a 1:1 basis, the turns and currents in the two coils are equal, and the figure may be regarded as merely showing a section through a single coil, of N z turns, or of Ar 2 /2*amp.-turns. The area of the core of this imaginary coil will be 6m 3 , where ra 3 is the mean circumference between the actual coils, and 6 is 204 ELECTRICAL ENGINEERING the distance between them. The flux produced in this region by the m.m.f., / 2 N 2 , is then o X 4-n A r 2 2 iw 3 , due to The number of interlinkages is This represents an inductance of L" 2 the secondary coil, since half of the inductance is due to the pri- mary and the other half due to the secondary. The total secondary inductance of this coil is then L 2 = Lff 2 = and the primary inductance is, similarly, T 27rAVr 9 c Li- j [2rai^ where c is the depth of the primary coil. Since ^ = N former, referred to the primary is L = ^y 1 -^!^ + If two legs are in series, L (tote/) or, if in parallel, L (tota0 = - In practical units, L = 32 X 10- 9 ~ 2> the total inductance on 1 leg of the trans- n. (105) 2L, m 2 henrys ' (106) where the dimensions are in inches. The same reasoning may be applied to a core-type transformer in which the coils are differently arranged, for example, as in Fig. 157. Here are two secondary coils, with the primary placed between them. Consider the primary as if made up of two equal coils, separated by a dividing line shown dotted. The calculation should then be made of the combined inductance of the secondary, S', and one-half of the primary, which are grouped MECHANICAL STRESSES IN TRANSFORMERS 205 together as A in the figure, and similarly, the secondary S" and the other half of the primary grouped as B. From (105), the inductance for A is, U and for B, in which ]2 |~,w,/ i m w'i = mean length of inside one-half primary turn w"i = mean length of outside one-half primary turn m' 2 = mean length of inside secondary turn m"z = mean length of outside secondary turn m 3 = mean length of inside gap ra 4 = mean length of outside gap w"i = 2rai m 3 + w 4 = 2m. If coils are symmetrical, mi m 2 . Supplying all of these values, the total inductance is L = L'+L" = -^r-\mi~ + m 2 % + ra&lcm., I L 6 3 J where Ni is the number of turns in half the primary coil. If TI is the number of primary turns per leg of the core, c a - + m 2 If dimensions are in inches, T 16 TYr c Lj -, /-vn T cm. per leg. mb] FIG. 158. In a similar manner, shell-type transformers may be dealt with. Such a transformer is shown in Fig. 158. In this, let m = mean length of 1 turn, NI = number turns in half of a primary 206 ELECTRICAL ENGINEERING coil, = one-quarter total primary turns. Using the same reasoning as with core-type transformers, the inductance of a unit combination, A, in the figure, is [| + I + b] cm. (107) Note that m ^ + + Z is the equivalent area, whence the total inductance is SL = j |g + F+&| cm. In inch units, 128 Calculation of Stresses. Under ordinary conditions of load, these would not be excessive, but for maximum current, as in the case of short-circuit, or heavy transient currents from switching, they may be very great. Calculation may properly be based on the short-circuit current, remembering again that a wire 1 cm. long, carrying 10 amp. (unit current), if placed perpendicular to a field of 1 line per sq. cm., is repelled by a force of 1 dyne; or the force in dynes = BI'l, where I' is expressed in absolute values abamperes. If the flux density in the gap, &, between coils, is B max then it may be assumed that the average density of the flux leaking D through the coils themselves is ^, which is then the average density of the flux passing through the coils of any section A, Fig. 158, and the force per turn on any coil, will be F t = -^ X 1\ X m dynes B max 98f grams ' where m is the mean length of the turn. If /2 is in amperes, B max z F t = grams per turn. Let the effective value of the short-circuit current be 7 2 , and let the total secondary turns be T 2 , then the turns in a half coil (Fig. 158) are - MECHANICAL STRESSES IN TRANSFORMERS 207 The maximum value of the force will be 2 X 9810 v 4 = 8X9810 m grams ( 108 > or in the case of the primary short-circuit current / m V2ITmB m 8 X 9810 where / is the effective value of the primary short-circuit current and T the total number of primary turns. The leakage flux must, in the case of short-circuit, be the main flux (neglecting the flux due to the voltage which is consumed by the ohmic drop), if it is assumed that the generating station is large and the voltage impressed upon the transformer is normal even though the transformer is short-circuited. (See note.) The maximum value of the flux between a group of coils is obtained by multiplying the maximum value of the flux density B m by the equivalent area as given in (107). That is The group contains in this case one-quarter of the turns and ET the voltage per group is -r where E is the effective value of the impressed e.m.f. The relation between the maximum value of the flux and the voltage is given by the well-known relation Substituting this in (109) (110) The average value of the force is obviously one-half of the maximum value. 208 ELECTRICAL ENGINEERING The force between the coils is proportional to the rating assuming the same regulation. NOTE. The actual flux enclosed by the secondary turns depends upon the terminal voltage and the ir drop. At short-circuit the secondary terminal voltage obviously is zero. Thus if as a limiting case the ir drop is neglected the secondary winding encloses no flux. As long as it is assumed that the primary voltage is normal voltage and that the ir drop is again neglected the primary coil encloses the same flux during the short-circuit as it does at no-load. The path of the flux must therefore be essentially different. In the latter case it traversed the two windings and is therefore mainly in the iron, while in the former case it must traverse only one winding the primary. Thus the flux must find its way between the primary and secondary coils and is thus the so-called leakage flux. CHAPTER XXXI GENERAL PRINCIPLES OF TRANSFORMER DESIGN Type. Transformers may be classified as belonging either to the "core type" or the " shell type." Core-type transformers frequently have a single magnetic circuit of rectangular form. On the two vertical sides of this core are placed the windings, each side being provided with half of the primary and half of the secondary coils, the low- voltage coils usually being placed next to the core (Fig. 159). Shell-type transformers usually have a multiple magnetic circuit the coils being placed upon a central core, the outer limbs of which extend around the coils, somewhat resembling a shell (Fig. 160). As illustrated diagrammatically in the figures, it is Secondary Core Primary FIG. 159. FIG. 160. seen that the coils of the core-type transformer have the form of a cylindrical shell, while those of the shell type are in the form of discs. The former lend themselves readily to designs of great mechanical strength, while the latter tend to be mechanically weak. The present tendency seems to be more and more toward the core type, and it remains for the superiority of the shell type to be demonstrated in any given case in order to justify its existence at all. Recently transformers having a multiple magnetic circuit have been introduced. The coils are of the cylindrical form placed around the central core. Thus, this is called the cruciform type. 14 209 210 ELECTRICAL ENGINEERING An important consideration with respect to the choice of type is the method of cooling the transformer. Core-type transform- ers are usually immersed in oil in such a way as to provide free circulation of the oil about all surfaces of the coils and core. The oil then receives the heat and carries it to the outside case which is frequently corrugated to present greater effective surface to the outer air. Shell-type transformers are cooled by the above method, but more frequently this is augmented by the addition of coils of pipe through which is forced a stream of cool water. These coils are placed in the oil above the transformer. The addition of the cooling water is essentially a feature of large transformers, since they have less area of possible cooling surface per unit volume than have smaller units. A common form of the shell type is known as the air-blast type. The method of cooling consists in forcing a continuous blast of cool air up through the ducts with which the core is provided, and between and around the coils. Efficiency. Transformers are not designed to give the highest possible efficiency as this would involve too great an expense in materials and manufacture, but, rather, the highest practical efficiency, so as to meet competition both in price and in quality. Consequently, from results obtained in practice, it is easy to construct a table of efficiencies which might reasonably be expected of various sizes of transformers of moderate voltages, say up to 10,000 volts. This table is as follows : Efficiency 25 cycles 60 cycles 1 94.0 96.0 5 96.5 97.5 10 97.0 98.0 50 98.0 98.5 200 98.0 98.5 Knowing the approximate efficiency of the transformer which is to be designed, the total losses are of course also known. For example, let it be required to design a 10-kw., 60-cycle, 200 %oo- v lt core-type lighting transformer. The efficiency is to be about 98 per cent. The losses are 2 per cent., or 0.02 X 10,000 = 200 watts. GENERAL PRINCIPLES OF TRANSFORMER DESIGN 211 Losses. These losses are made up of the I 2 r loss in the copper windings and the hysteresis and eddy current losses in the iron core and windings. Maximum efficiency is obtained at that load for which the copper and iron losses are equal. It becomes a matter of choice in design as to what ratio shall be given these losses or at what load they shall be equal. Thus, for power purposes, the copper and iron losses should be about equal at full-load, giving maxi- mum efficiency at full-load. For lighting purposes, however, owing to the peculiar conditions of operation, this is not generally desirable. A lighting transformer carries full-load only for a very small period during each 24 hr., while the rest of the time it is operating practically at no-load. Thus the copper loss is quite small even with a large value of 7 2 r, while the core loss is larger since it is continuous through the whole day. It would be better, therefore, to make the copper loss relatively greater than the core loss, at full-load, and thus reduce the total losses for the daily operation. Fairly good values to choose for these losses are: copper loss = 60 per cent., core loss = 40 per cent, of the total loss. In the example considered, copper loss = Pr = 200 X 0.60 = 120 watts, core loss = 200 X 0.40 = 80 watts. The core loss may be further divided between loss due to hysteresis and loss due to eddy currents. The former is usually larger because it depends on the magnetic quality of the iron or steel used, whereas the latter depends largely on the degree of thinness of the laminations of the core, and this may be carried to any extent mechanically practical. Values of hysteresis and eddy current losses when silicon steel laminations .014 in. thick are used are: hysteresis loss = 0.7 watt per Ib. at 60 cycles, eddy current loss = 0.3 watt per Ib. at 60 cycles, when the maximum induction density is 64,500 lines per sq. in. (10,000 lines per sq. cm.). Since 1 cu. in. of this material weighs 0.28 Ib., the loss per cu. in. at 60 cycles and 64,500 lines per sq. in. is: hysteresis loss per cu. in. = 0.28 X 0.7 = 0.196 watt, eddy current loss per cu. in. = 0.28 X 0.3 = 0.084 watt, total core loss per cu. in. = 0.28 watt. 212 ELECTRICAL ENGINEERING Hysteresis loss for any frequency and density is given ap- proximately by the equation, hyst. loss = W h = 0.196 X ^ X where V = volume of iron. Similarly, eddy current loss is / V, 0.084 X From these two equations and the core loss which is given, the volume may be obtained for any value of B. Assuming, as will later be done, that B = 70,000, in the example, * 80 ^ [0.196 X (1.086) 1 - 6 + 0.084(1. 086) 2 ] = 80 0.2205 + 0.099 = 250 cu. in. And the hysteresis loss is W h = 0.196 X 1.125 X 250 = 55.2 watts, and the eddy current loss is W e = 0.099 X 250 = 24.8 watts. 5 10 15 20 Volume per Watt Hysteresis Loss Cu, In. 0.1 0.2 0.3 0.4 0.5 0.6 Watts per Cubic Inch FIG. 161. B and V. The relation between B and V is shown by the following curve, Fig. 161, from which it is evident that values of B should lie between 50,000 and 90,000. GENERAL PRINCIPLES OF TRANSFORMER DESIGN 213 Hysteresis and Eddy Current Loss per Cubic Inch. (!)/ = 60 B 50,000 60,000 70,000 80,000 90,000 100000 B 775 93 1 085 1 24 1 395 i fit) 64,500 / B \i.e Of\K OQQ 1 19^ 1 d.9 Iijt* 2f\f \64,500/ TF OIQfi ( B V 6 0197 01 79 f\ 099O 097ft . to 00.^0 . uo OAfyy W k - 0.19G ^ 4 1 B y OAfll OftAC 11 ft .^/o 1A .o^o IQr .4UZ 241 \64,500/ TF n ns4 / B \ 2 .OU1 OfkKfJK .oDO OO797 . lo Onoo . O4 01 9QJ. .O 01 AQQ .41 Oonne W. - 0.084 ^ 64j50Q j TF 7 0.1775 0.2452 0.319 0.4074 . iDoy 0.5069 0.6045 (2) / = 25; = 0.417; = 0.174 W K 0.053 0.072 0.0917 0.116 0.143 0.1675 W, W 0.0088 0.0618 0.0127 0.0847 0.0172 0.1089 0.0225 0.1385 0.0285 0.1715 0.0384 0.2059 v As a matter of fact the usual limits are: for 60 cycles, B lies between 60,000 and 75,000, for 25 cycles, B lies between 80,000 and 90,000. In the example, let B = 70,000, which will be taken as a trial value. From Fig. 162 the volume per watt loss by hysteresis is 4.55 cu. in. The total volume of iron is 4.55 X 55.2 watts = 250 cu. in. Magnetizing Current. Having chosen a suitable value of B, we can at once find out the required number of ampere-turns per inch length of magnetic circuit, from the saturation curve, p. Let M o = ampere-turns per inch and I = length of magnetic circuit. Then total ampere-turns = M Q l = -\/2imt, where -\/2i m = maximum value of magnetizing current and t = number of turns on the primary. Using the fundamental equation for e.m.f., E = 4.44 ft* X 10- 8 = 4A4ftBA X 10~ 8 , 214 ELECTRICAL ENGINEERING the magnetizing volt-amperes are 1-1 " X 10 8 = vfBAMol X 10~ 8 = irfBMoV X 10~ 8 , since 4.44 = \/2ir, and V=Al. The percentage magnetizing current is obtained by dividing by El, thus, T = 10 u Xkw.' In the example, M is found to be 6.5. Therefore, substituting known values into the equation, i m 3.14 X 60 X 70,000 X 6.5 X 250 T : io" x 10 l5 ' or approximately 2 per cent. This is a reasonable value. In practice, magnetizing currents range from 2 to 8 per cent., being larger in smaller transformers and at lower frequencies. Number of Turns, Total Flux, Area, and Length of Magnetic Circuit. Returning to the fundamental e.m.f. equation, it is seen that turns and flux are both unknown. A practical limit in help- ing to decide what value to assign to either one of these unknowns is found from the fact that the number of turns should depend upon the voltage. While it would not be safe to allow too great a difference of potential to exist between adjacent turns, this consideration is not the deciding feature. The choice of number of turns is governed largely by cost considerations. From prac- tice it is known that volts per turn should lie between 0.4 X \/kw. and 0.6 X A/kw. in core-type transformers. The former value is more suitable for distribution transformers when it is desirable to keep down the core loss, while the latter is suitable for power transformers. The value for shell type is from two to three times as great. In the example, it will be assumed that volts per turn = 0.5 X = 1.56. Then, turns on primary GENERAL PRINCIPLES OF TRANSFORMER DESIGN 215 and flux - 586 > 000 ' area $ 586,000 = A = B =: "TpOO" == 8 ' 37 sq ' and length, Resistance, Length of Mean Turn, Total Length and Size of Windings. Returning now to the windings, it is possible at first to calculate the primary resistance, since the copper loss and the current are known. In the example Pr =120 watts. This must be divided be- tween primary and secondary, and half may be assigned to each, as a reasonable approximation. Thus primary Also, Pr = = 60 watts. W 10,000 7l = Yi = " "2000T = 5 60 .'. #1 = ^ = 2.4 ohms. &o Knowing the resistance and number of turns, the size of wire may be found when the mean length of one turn is estimated. As a basis for this, the cross-sectional area, A, of the core is known, and experience tells about how much space is necessary for insu- lation between core and coils and for circulation of the cooling oil between the coils. Also, since the heat generated in the in- terior of the coils has to pass through the thickness of copper and insulation, it will be unwise to make the coils too thick. Practical thickness of insulation against voltage is given in the following table. TABLE VII ,. ,. Insulation Volta thickness (mils) 110 40 440 50 1,000 70 2,300 100 6,600 180 16,000 260 216 ELECTRICAL ENGINEERING For circulation of oil, space of not less than J m width should be allowed. This width is governed by the height of the coils. Thickness of the coils should hardly exceed 1 in., but may reasonably be % in. Applying this procedure to the example, it is found that with an area, A = 8.37 sq. in. of iron, the gross area occupied by the laminations will be about If this area is in the form of a square, the side of the square will be -\/9l3 = 3.05 in. Fig. 162 is next drawn, showing the relative positions of coils, core, insulation, etc. In this case, the length of mean turn of the secondary winding is L 2 = 4 X 3.05 + 2^(0.25 + 0.04 + 0.375) = 16.4 in. Since the secondary winding is nearest the core, its features will be discussed first, thus avoiding any error in the final determination of TI ,~\ t75- ^ V -f -1 1 25, = 3.05 :=> Core ^ 1 Primary FIG. 162. the mean length of primary turn. Total length of secondary is 16 4 X < 2 = -TH- X 12 ft. In general, 100 2000 X 1280 = 64. In practice, however, it is found convenient to put the two primary coils in series and the two secondary coils in parallel to obtain the 20: 1 ratio. If this procedure is adopted, the voltage impressed on one , 2000 primary coil is -y- = 1000, while the whole secondary voltage of 100 will be across each of the secondary coils. The secondary turns per coil will then be 100 1000 X 640 = 64 GENERAL PRINCIPLES OF TRANSFORMER DESIGN 217 and each coil will carry half of the total secondary current or 10,000 = 50 amp. Total length of each secondary coil is then L 2 X t, = ~ X 64 = 87.3 ft. 27? Resistance per 1000 ft. of secondary coil = AQ i * U.Uo7o Since the coils are connected in parallel, the resistance of one coil is 2R Z . The resistance of a secondary coil is obtained from the fact 60 that the secondary copper loss per coil is -^ = 30 watts. Thus, the resistance of each secondary coil is 30 = - 012 onm - Resistance per 1000 ft. of conductor is n ' 7Q = 0.1375 ohm. U.Uo/o This corresponds to an area of 0.07 sq. in. The conductor chosen must be of copper strip, of rectangular cross-section. In using strip, the practical dimensional limits are about 0.1 in. in thickness and 0.5 in. in width. These dimensions give an area of 0.05 sq. in. If greater area is re- quired, any number of strips may be wound in parallel. Each strip is, however, insulated, usually with double cotton covering, to prevent too great eddy current loss in the copper. In the present case there will be two strips required, each of 0.1 X 0.35-in. section. With insulation, the dimensions of the double conductor become 0.36 X 0.22 in. It will be seen that the most practical arrangement of the turns will be to have two layers deep and 32 turns per layer. Then the thickness of the coil becomes 2 X 0.22 in. = 0.44 in.; the length of the coil is 32 X 0.36 in. X 11.5 in. The corrected mean length of turn is L 2 = 4 X 3.05 + 27r (0.25 + 0.04 + 0.22) = 15.4 in. I K A y f\A Total length = - ^~ - 82 ft. Resistance per 1000 ft. = ^ = 0.147 ohm. 1375 Corrected cross-section is X 0.07 sq. in. = 0.0655 sq. in. 218 ELECTRICAL ENGINEERING Maintaining the same thickness, i.e., 0.1 in. the width of the strip, with insulation, now becomes 0.332 in., and the coil length is 32 X 0.332 = 10.6 in. The mean length of the primary turn may now be found. It is L l = 4 X 3.05 + 2T (0.25 + 0.04 + 0.44 + 0.04 + 0.25 + 0.1 + 0.375) = 12.2 + 2w X 1.495 = 21.6 in. Total length of primary is then / 21.6 X || = 21.6 X = 2304 ft. r> . 24- Resistance per 1000 ft. of primary is 2304 = 2394. = ohms. Referring to wire tables, this resistance is found to be nearly that of No. 10 B. & S., which has resistance of 1.18 ohms per 1000 ft. at 65C. If now it should be desirable to use copper strip for the primary winding, the requisite area may be found by comparison with that of No. 10 wire. Thus, 1 18 area = - X 0.00815 = 0.00922 sq. in. In this case, however, it will be practical to use No. 10 wire. Wire larger than No. 10 is not generally used, but smaller sizes are preferable to rectangular strip. Choosing then No. 10 wire the space which the 640 turns of each coil will occupy must be determined. With a layer 10.3 in. long there will be 10 3 Q IQ2 = 100 turns per layer, and 640 100 =: 6 ' 4 layers ' Obviously, the best arrangement of these 640 turns will be to have 8 layers of 80 turns each, giving a coil length of 8.24 in., and coil thickness of 0.824 in. The thickness will be slightly less, owing to the bedding of the layers. Perfect bedding would give 0.824 X 0.866 = 0.714 in. The value of 0.75 in. originally assumed may therefore conveniently be taken as correct. The mean length of the primary turn is then LI = 21.6 in. as previously calculated. GENERAL PRINCIPLES OF TRANSFORMER DESIGN 219 21 6 Total length of primary = -^ X 1280 = 2300 ft. Total primary resistance = 2300 X 1.18 = 2.72 ohms. This resistance deviates considerably from the value of 2.4 ohms assumed, but not enough to warrant the choice of another size of wire for the primary. Having determined the core cross-section, and the coils, an assembly sketch may be made, as shown in Fig. 163. Allowing 0.3 in. between the coils on the two legs, the size of the window is FIG. 163. found to be 4.24 in. wide by 10.75 in. high. The total core height is 10.75 + 3.05 + 3.05 = 16.85 in.; total core width is 4.24 + 3.05 + 3.05 = 10.34 in. The total volume of iron is length X net cross-section, or, V = (2 X 16.85 -f 2 X 4.24) X 8.37 = 42.18 X 8.37 = 354 cu in., which does not compare very favorably with the first assumption of 250 cu. in., but this is not very important since it should be noted that I calculated from core loss and I calculated from 220 ELECTRICAL ENGINEERING the space required for the copper windings will generally not be in agreement. We must have sufficient space for the windings, but I should not be any greater than necessary. Therefore, unless we wish an entire recalculation of the design based on altered assumptions it is sufficient to accept the new value of I and the attendant new value of V. The mean length of the flux path is, I = 2 (10.75 + 4.24) + 27r X 1.5 = 29.98 + 9-44 = 39.42 in., as against 29.5 in. in the preliminary calculations. Per Cent. Magnetizing Current and Core Loss. Applying the new values of V and I, we may obtain new values for per cent. magnetizing current and the core losses. Thus, amp. turns = M Q X I = 6.5 X 39.42 = 256. 256 Max. exciting current = \/2i m = TQC = 0.2. ' Per cent. i m = = = 0.028. /i 5 Hysteresis loss, W h , will be in the ratio of the two volumes thus far obtained, namely; 250 cu. in. and 354 cu. in. 354 Wh = 250 x 55<2 = 78 * 2 watts ' and similarlv the edd y 354 current loss is W e = X 24.8 = 35.1 watts, giving a total core loss of 113.3 watts. Efficiency. The approximate efficiency is then input losses 77 = -- ; - > input where the losses are: Primary copper loss = 5 2 X 2.72 =68 watts. Secondary copper loss = 2 X 50 2 X 0.012 = 60 watts. Core loss = 113.3 watts. Total loss = 241.4 watts. *'* * = ~ 10 ooo = * 976 = 97>6 per cent> It is seen that the efficiency is very nearly that which was assumed at the outset, so that the variations in values, even where GENERAL PRINCIPLES OF TRANSFORMER DESIGN 221 they have been large, as with volumes and length, have not been such as to produce any considerable effect. TT copper loss . However, the ratio of ^ ron | oss nas decreased, the former now constituting only 53 per cent, of the total loss, instead of 60 per cent, as at first assumed. There is, of course, nothing hard and fast about these relations as it is impossible to say just how the transformer will be oper- ated from day to day. If it is desired to approximate more closely to the original assumptions of losses and efficiency, it will be necessary to go back to the beginning and choose from among the numerous variables, let us say, another value of turns and another value of the flux density. However, in the present instance, the values so far obtained will be regarded as satisfactory, and then there remains only to determine the regulation, heating and cost of material, to see if these also will be satisfactory. Regulation. In Chap. XXVI, p. 182, the regulation of a trans- former was found to be Reg. = ^ " 1> where EI = V E 2 2 + 2E z (i 2 r Q + 0.54n> - i'&o + 0.5i m z ) approximately. Of these quantities, EI = 2000 volts, assumed impressed on the primary, i z = 7 2 = 5 amp. = the energy component of the load at unity power factor (assumed) and referred to the primary basis. On this assumption, the wattless component of the load current, i'z = 0. 4 = energy component of the exciting current. To obtain 4, we have: core loss = Eih = 113.3 watts. 1100 " ih = ^ = - 0566 amp *' and too = Vi^+4 2 = V(H41 2 + 0.0566 2 = 0.152 amp. r = fli + # 2 = 2.72 + 0.009 X 400 = 2.72 + 3.6 = 6.32. To determine X Q , the combined leakage reactance of primary and secondary, we have X Q = 27T/ (Li + 1/2) = 27T/I/0. LI and L 2 may now be calculated by the help of equations, p. 204, Chap. XXX, but each must be done separately since the 222 ELECTRICAL ENGINEERING two primary coils are in series while the secondary coils are in parallel. We have Referring to Fig. 155, the constants in these equations are readily evaluated. Thus, we have, Ni = 640; N* = 64; I = 10.75, mi = mean length of primary turn = 21.6 in., mz = mean length of secondary turn = 15.4 in., mz = mean length of gap between coils = 18 in., a = secondary coil thickness = 0.44 in., b = distance between coils = 0.39 in., c = primary coil thickness = 0.75 in. Supplying these values, fc - 64 X 10-9 41 X 1Q4 2L6X ' 75 18X ' 39 Ll ~ 10.75 3 = 0.00244 [5.4 + 3.51] = 0.02175 henry = 6.1 X 10-6 [2.26 + 3.51] = 0.0000352 henry, where L 2 is the actual secondary inductance. Referred to the primary, L 2 = 0.0000352 X 400 = 0.0141 henry. Lo = Li + L 2 = 0.02175 + 0.0141 = 0.03588 henry and Zo =27r/Lo = 377 X 0.03588 = 13.53 ohms. Supplying all the values into the formula for Ei, we have, Ei = 2000 = X 6.32+0.5 X 0.0566 X 6.32 + 0.5 X 0.141 X 13.53) 4 X 10 6 = # 2 2 + 2# 2 (31.6 + 0.179 + 0.95) = V 2000 .'. E 2 = 1968 volts, and regulation 1 1 = 1.016 1 = 0.016 = 1.6 per cent, for full non-inductive load. Heating. The total radiating surface of each primary coil is found by calculation to be 388 sq. in. Therefore, the watts per square inch which must be radiated from the primary coil are . . 0875 . 388 sq. in. GENERAL PRINCIPLES OF TRANSFORMER DESIGN 223 Similarly, the area of the secondary coil radiating surface is 340 sq. in. The watts per square inch that must be radiated are - The radiating surface of the core is about 400 sq. in. Therefore watts per square inch that must be radiated are __ 113.3 watts ' 400 sq. in. Watts per square inch serve as an empirical guide by which it may be determined satisfactorily whether the design is sufficiently liberal to permit of dissipation of the heat without undue rise of temperature of any part. In general a loss of 0.4 watt per sq. in. of surface of the coils and core is quite satisfactory. In designing the case, however, about 0.15 watts per sq. in. only should be allowed. In the transformer, then, since the entire loss in watts must be radiated from the case we should need an area of 241.3 watts . . .,, ,, ., fT-^= = 1610 sq. m., in contact with the oil. U.lo Weight and Cost of Material. The core volume has been found to be 354 cu. in. At 0.28 Ib. per cu. in. the core weight is 354 X 0.28 = 99 Ib. Cost of core at 3.5 c. per Ib. is 99 X 0.035 = $3.46. The primary copper volume is length X section, = 2300 X 12 X 0.00815 = 225 cu. in. Secondary copper volume is 82 X 12 X 0.0655 X 2 = 129 cu. in. Total volume of copper is 225 + 129 = 354 cu. in. Weight of copper at 0.32 Ib. per cu. in. is 354 X 0.32 = 113.4 Ib. Cost of copper at 16c. per Ib. is 113.4 X 0.16 = $18.15. Total cost of iron and copper is $3.46 + $18.15 = $21.61 Of course, such a calculation of cost has comparative merit only, as it does not include labor or such materials as insulation, oil and case. 224 ELECTRICAL ENGINEERING Summary of data of 10-kw., 60-cycle 2000-100-volt core-type distributing transformer. - High side Low side Kilowatts 10 Frequency 60 Ratio of transformation 20:1 Volts 2 000 100 Arnp6r6s 5 100 Window dimensions, in 10% by 424 Total width of iron!! in Total height of iron, in 10.34 16.85 Depth of lamination, in Electrical Number turns in series 3.05 1 280 64 Section of conductor Amperes per square inch 0.00815 614 . 0655 763 Number of coils 2 2 Connection of coils. ... Series Parallel Width of coil 75 44 Height of coil 8 24 10.6 Number turns per coil 640 64 Mean length of turn Resistance of circuit at 65C 21.6 2 72 15.4 0.009 Magnetic Total maximum flux Effective core section, sq. in. 586,000 8 37 Effective core length, in Core density Effective core ampere-turns 39.42 70,000 180.5 Magnetizing current Thermal /'flloss 0.141 68 60 Radiating surface of coil 388 340 Watts per square inch 0875 0885 Core loss 113 3 Radiating surface of core Watts per square inch 400 28 Total loss, full-load 241 3 External radiating surface Watts per square inch 1,610 15 Efficiency and regulation Per cent, core loss, full-load Per cent, copper loss, full-load . 1.15 1 3 Per cent, efficiency, full load 97 6 GENERAL PRINCIPLES OF TRANSFORMER DESIGN 225 High side Low side Per cent, magnetizing current Per cent, resistance 2.8 1 58 Per cent, reactance. . . 3 38 Per cent, regulation 1 6 Weight and cost Copper, pounds 113 4 Iron, pounds 99 Pounds copper per kilowatt 11 34 Pounds iron per kilowatt . 9 9 Cost of copper at 16c Cost of iron at 3.5c Total cost. $18.15 $3.46 . $21 61 Cost per kilowatt $2 16 Having now developed the general principles and procedure in transformer design, it is desirable that the student should carry through the calculations for some assigned machines. Trans- formers of different capacity may be assigned to the students of a section, each student being required to complete the calculations for both 60 cycles and 25 cycles, tabulating the specifications and making sketches to scale of the core and windings. Many of the finer points in design are omitted here, since the principles are the primary interest. For practical designing the fact that experience is a factor of the greatest importance should always be remembered by the student who is attempting to master the practical aspects of the subject. To assist in the further study of the principles of transformer design, the following suggestive questions are added. 1. Find regulation at 80 per cent, power factor. 2. Why are less volts per turn used with lighting than with power transformers? 3. If the core loss is too great, how may it be reduced? 4. What relation does per cent, exciting current have to core loss, copper loss, efficiency and regulation? 5. How may per cent, exciting current be reduced? 15 CHAPTER XXXII COMBINATIONS IN MULTIPHASE TRANSFORMER SYSTEMS When the primary of a transformer is connected to a source of e.m.f . the following equation relating the impressed e.m.f . cur- rent, resistance and inductance obviously obtains e = ri + ^ (Li). The drop, ri, is small, being perhaps 5 per cent, of 1 per cent. of the normal voltage, if the exciting current is 5 per cent, of normal current and the resistance drop at full-load is 1 per cent. Thus the e.m.f. consumed by the transformer counter e.m.f. is approximately If the transformer were merely a coil having an air core, the inductance would be constant, and the induced voltage would be . di If i were a sine wave of current, e would also be a sine wave displaced 90 behind i. However, with iron cores, as with trans- formers, the inductance is not constant, but is a function of the current i. Hence di . . and if i is a sine wave of current, the counter e.m.f. of self-induc- tion is no longer a sine wave. Similarly, if a sine wave e.m.f. is impressed on the transformer, the current will not have the sine shape, but will be made up of fundamental, 3d, 5th, etc., harmonics. In Chap. XXVIII, the characteristic wave of exciting current with a sine wave e.m.f. impressed was determined. On analyzing this wave by FOURIER'S series as indicated in problem 78, it is found to consist of a fundamental and triple with higher har- 226 MULTIPHASE TRANSFORMER SYSTEMS 227 monies of lesser amplitudes. The presence of the triple-fre- quency wave is an important feature in the exciting current of every iron-cored transformer operating with impressed sine wave e.m.f. If, however, the triple frequency wave is suppressed in some way, as is often the case in three-phase systems, so that the ex- citing current is of sine shape (neglecting small higher harmonics), the induced voltage, and consequently the terminal voltage, will not be of sine shape, but will have the characteristic form shown on p. 194. The transformer, of course, must generate its' own counter e.m.f. or the induced voltage, and hence may be regarded as a generator, electrical energy being supplied to it instead of me- chanical energy. So far the transformer has been dealt with as a single unit. It is common practice, however, to group transformer units in various ways so that they shall serve as group units in the trans- mission and distribution of energy in systems other than the single-phase system. The Three-phase System. While two-phase, four-phase and six-phase systems are used to some extent and under certain con- ditions, yet the three-phase ^ system in its various forms is far more important than all of these. Its study forms a basis for the development of any multiphase theory. The principles of two-phase and three-phase working from the standpoint of the alternator are explained in Chap. XXXV. In the present in- stance, three-phase will be dealt with in reference to the transformer alone. Consider three similar transformers, A, B and C, receiving current from three sources of simple sine waves of e.m.f. (Fig. 164). Let the voltage impressed on A be e t = EI sin 6, that on B, e* = E l sin (0 + 120), that on C, e 3 = EI sin (6 + 240). In each case the arrows in the figure indicate outgoing and return wires. FIG. 164. 228 ELECTRICAL ENGINEERING Neglecting higher harmonics the currents flowing will be, re- spectively ii = /i sin (0 0) i z = I, gin (6 + 120 - 0) i z = /! sin (0 + 240 - 0). If the transformers are so arranged that the return currents shall flow through the same wire, as in Fig. 165, the value of the current in this return FIG. 165. wire will be i n = ii + iz + iz = Ii (sin (0 - 0) -f sin (0 + 120 - 0) + sin (6 + 240 - 0). The student should prove that this current is zero. He should prove, also, that if there is current of triple frequency flowing in the lines, the triple frequency current in the fourth or neutral wire will be three times that in any line. Problem 79. Let the three line currents be given by the equations: ii Ii sin (0 $0 + /a sin (30 that is, the current in the neutral wire is three times the sum of all odd harmonics which are multiples of three which are present in any line wire, all other harmonics becoming zero in the neutral. Voltage Waves in Three-phase, Four-wire System. The neu- tral wire serves to make the system virtually three single-phase systems instead of a three-phase system having peculiarities of its own. Thus if a sine wave e.m.f. is impressed on each trans- former, the e.m.f. between lines is the vector sum of any two of these and is also a sine wave. Since there is no current of fundamental frequency in the neutral wire, there is no necessity of having the wire there. Its absence will not be the occasion for any interruption in the circuit. However, since the circuit of the higher harmonics has been inter- rupted, it becomes of importance to study higher harmonic effects in connection with three-phase and with any other systems. Transformers so arranged with or without the neutral wire are MULTIPHASE TRANSFORMER SYSTEMS 229 said to be Y-connected. If the circuits were unbalanced the situation would be somewhat different, as will be discussed later. For the present, however, it is sufficient to see that a three-phase circuit may be composed of only three wires, each representing the outgoing wire of one of the phases. Three-phase, Y-connected Transformers. Let it be assumed that the three, so-called, phase voltages are OA = BI = E l sin B + Es sin (30 + a), OB = 6 2 = E l sin (0 + 120) + E z sin (3[0 + 120] + a), OC = e 3 = #1 sin (0 + 240) + E 3 sin (3[0 + 240] + a), these voltages being represented vectorially in Fig. 166. To find the line voltage AB. Evi- dently this is e AB = AO + OB = - since it is taken in direction from A to B. Directions from outward are taken as posi- tive. Therefore CAB = - EI sin - E z sin (30 + ) + E l sin (0 + 120) -f E 3 sin (3[0 + 120] + a) = #i[sin cos 120 + cos sin 120 - sin 0] + Et[sm (30 + a) - sin (30 + a)]. The last term vanishes since sin (3[0 + 120] + a) = sin (30 -f 360 + a) = sin (30 + a) and e AB = 1.73 EI sin (0 + 150). The student should prove this by performing the intermediate operations. Thus, it is seen that in a balanced three-phase Y-connected sys- tem, a triple frequency e.m.f. cannot exist in the voltage between the lines. The same will be shown to be true also for what is called the A-connection. This does not mean, as stated, that there can be no triple frequency e.m.fs. in the phase windings, but simply that they cannot be between the lines. Likewise, the other line voltages are: e BC = - 1.73 EI sin (0 + 90) CCA = 1-73 E! sin (0 + 30). These voltages are represented in Fig. 167 (a) and (6), which give two ways of representing the same thing. The line and phase voltage relations may also be shown graphically, as in 230 ELECTRICAL ENGINEERING 167. .e Fig. 168. Here, as in the equation for e^ B) OB is combined with OA reversed. Carrying out the same process by which the assumed triple- frequency voltages in the line were eliminated, it could also be found that any higher harmonics which were multiples of 3, as 9th, 15th, 21st, etc., would vanish. Even harmonics are of necessity absent if the waves are sym- metrical. Thus there remain only the 5th, 7th, llth, 13th, 17th, 19th, etc., which could exist in the lines. In the three-phase Y-connection, the triple frequency voltages in the phases are all in time-phase with each other, and the phase therefore acts like three circuits in parallel. Their extremities could be joined without causing any triple frequency current to flow. If the neutral point, 0, be connected to ground, the triple frequency in the phases would cause all three transmission lines to oscillate just as would be the case with a single-phase line one side of which was grounded. In this case the three lines cor- respond to the ungrounded side of the single-phase line. Three-phase A-connected Transformers. W hen three transformers are so connected as to form a closed circuit, there are two facts in connection with their operation which are of great interest, namely: (1) there can be no circulating cur- rent of fundamental frequency in the windings; and (2) there always flows in the windings a current of triple frequency or an odd multiple of triple frequency. In proof of the first fact let the phase voltages be assumed, as before, Ci = EI sin + E z sin (3 + a) e 2 = E l sin (0 + 120) + # 3 sin [3(0 + 120) + a] e 3 = Ei sin (0 + 240) + E* sin [3(0 + 240) + a)] Adding the fundamental components, E l [sin 6 + sin ( 0+ 120) + sin (0 + 240)] FIG. 168. MULTIPHASE TRANSFORMER SYSTEMS 231 FIG. 169. = Ei [sin + sin cos 120 + cos sin 120 + sin cos 240 + cos sin 240] = Ei [sin + sin X (- 0.5) + cos X (0.866) + sin X (- 0.5) + cos X (- 0.866)] = EI [sin sin 0) = 0. Thus, if there is no e.m.f. of fundamental frequency acting in the closed winding, there can be no current of fundamental frequency circulating in it. In proof of the second fact, adding the triple-frequency com- ponents gives # 3 sin (30 + a) -f E z sin [3(0 + 120) + a] + #3 sin [3(0 + 240) + a) = SE's sin (30 + a). Thus if the delta is open at one point (Fig. 169) the triple voltage across the opening is three times the triple-frequency voltage of one phase. Similarly, with a Y-connec- tion with neutral point grounded, the triple-fre- quency current flowing into the ground is three times the triple-frequency current of one phase. When the neutral is grounded, it is no longer necessary to regard the system as three-phase, but it may be considered as three single phases having a common return, just as with the three-phase four- wire system already discussed. The triple-frequency current is then perfectly free to flow in each line wire, returning by way of the neutral, whereas without the neutral, the triple-frequency current cannot exist. To find the relation between line current and phase current in a A-connected system. Let the direction of the phase currents be assumed as indicated in Fig. 170 where ii = 7i sin + 7 3 sin (30 + a) iz = h sin (0 + 120) + 7 3 sin [3(0 + 120) + a] i 3 = 1 1 sin (0 + 240) + 7 3 sin [3(0 + 240) -f a] Taking the direction of the arrows as positive, the funda- mental line current, *A = ii ~ i* = 1 1 (sin - sin (0 + 120)) = 1.73/isin (0 - 30) FIG. 170. 232 ELECTRICAL ENGINEERING This relationship is similar to that of the voltages for Y-connec- tion. Similarly, also, there can be no triple-frequency current in the line. Voltage Waves with Y-connected Transformers. In problem 76, was assumed a sine wave of exciting current. This is approx- imately the case with the three-phase Y-connection since there can be no triple-frequency current in the line or phase. The phase voltage must then look like that of Fig. 151. The line voltage as previously seen will be a combination of two phase voltages, one of which is reversed, as in Fig. 171. (The depression in the line volt- age wave is not actually as deep as would appear from using sine waves of magnetizing current.) If the generator develops a sine wave e.m.f . and the transformer counter e.m.f. is much distorted, due to the hysteresis loop effect, then the difference between these two waves must be taken up by drops along the lines and in the apparatus. Problem 80. Given three A-connected transformers. Make a picture of the e.m.f. and compare it with that of a single-phase circuit. What is the shape of the wave of phase current? What is the shape of the wave of line current? Show also, by a sketch that the sum of two exciting current waves of a three-phase A-connected system makes nearly a sine wave, the triple frequency vanishing. How does the core loss in this system compare with that of three single-phase circuits? Problem 81. Given three Y-connected transformers. The line current can contain no triple harmonics but only 1st, 5th, 7th, llth, etc., harmonics. The phase voltage, however, has a large triple harmonic. The line voltage is a combination of two-phase voltages. What is the ratio of the phase voltage to the line voltage? Is it 58 per cent.? Evidently it is higher, as the phase voltage contains also the triple- frequency voltage. How does the core loss of this system compare with that of the A- connected system and with the single-phase? The flux wave is flat, since there is no triple-frequency current. Therefore the p IGi 172. maximum value of flux is less, and the core loss is less (by about 30 per cent.), than with sine waves of flux. Moreover, the exciting current is less because the max. value of the flux density is less. With open delta connection what will the voltmeter read? Evidently three times the triple-frequency voltage per phase. Why? With the MULTIPHASE TRANSFORMER SYSTEMS 233 neutral wire connected in a Y-system as in Fig. 172, what would be the current in the neutral? As has been pointed out, this is no longer a real three-phase system, but three single phases in which the neutral wire is common to all the phases. Evidently the triple-frequency currents can flow in each phase, and since they are all in time-phase with each other the neutral will carry three times the triple harmonic current of each phase. What then, will be the effect on the core loss of connecting in the neutral, as compared with leaving it out? These problems are stated in such a way as to form the basis for a fairly complete discussion, on the part of the student, of the effects which would be produced by the different ways of connecting the transformer. Three-phase Transformers. A natural development in the use of three single-phase transformers for three-phase work is the substitution therefor of a single three-phase transformer. (d) Po Po ' p o (e) Po FIG. 173. Let there be three cores of laminated iron, symmetrically placed, connected by legs, each core having on it the windings of one phase. This may be done as in Fig. 173, a, b. How, then, should the sectional area of the core be calculated? This should evi- dently be done in the regular way since each leg has its coil, and must have its flux set up by the coil. The yoke, however, cor- responds to a A-connection, and the flux in any leg of the yoke is = = 58 per cent, of the flux in any core. The yoke may also be formed as a Y-connection (Fig. 173, c, 6), in which the flux in any branch of the Y is the same as that in any leg. 234 ELECTRICAL ENGINEERING In practice, however, it is common to employ a form such as Fig. 173, d y in which there are three equal legs carrying the coils and the yoke is straight across the top and bottom. The whole core is built up of laminations, which, except for the diffi- culty of placing the coils, could be of one piece. In Fig. 173, d, the flux paths are outlined by dotted lines, and it is evident that so far as the magnetic core is concerned, coils 1 and 3 are sym- metrical with respect to each other, while coil 2 is unsymmetrical with respect to 1 and 3. Assume the reluctance of one leg to be p, and the reluctance of one section of the yoke to be po. Then there may be constructed an analo- gous electric circuit, as in Fig. 173, e. This gives the magnetic circuit which is supplied with flux by the m.m.f. of coil 1, on leg 1. It may be simplified to Fig. 174, in which . p' p p + 2p FIG. 174. p' = p(p 2p ) 2(p + po) The total reluctance of the magnetic circuit of coil 1, is thus 2p (3p -f 2 Po ) ^- 2P: "2(p + It is also evident that pi = p 3 . For p 2 , the circuit may be considered as made of two parallel paths, as in Fig. 175. Here, 1 p ; P2 = 2 (^P + 2p ). FIG. 175. Thus the relative reluctances of the two circuits are P2 Pi PO 2p In a good transformer, the ratio of height to width of the window is from 4 to 8:1; the average is about 6:1. .'. P is from 4 to 8 times as large as p . MULTIPHASE TRANSFORMER SYSTEMS 235 Assuming p = 6p , P2 _ Tpo _ 7 Pi 8po 8 .'. p 2 has 87K per cent, as great a value as Pl , and the exciting current in the middle coil is from 80 per cent, to 90 per cent, of that in the outside coils. Suppose it were necessary to have equal exciting current in all the coils. This could be accomplished by reducing the section of the middle leg. But, in this case, the core loss would be unbalanced, for it is approximately proportional to the square of the flux density which would be increased in the middle leg. Therefore, there must be some unbalancing. In practice all parts are made of equal section, including the yoke. It is there- fore easy to calculate the saving in material over three single- phase transformers. Question. Considering wave shapes as discussed above if the coils are connected Y, will there be a triple-frequency voltage? It has been shown that the triple-fre- quency currents are in time-phase in the different phases. Hence they produce fluxes in time-phase with each other. These then neutralize each other or pass around through the air which makes them very weak. The induced triple-frequency volt- ages are therefore very small. Thus a three-phase Y-connected transformer acts like three choking coils as far as the triple-fre- quency current is concerned. Their flux paths being largely in air, the hysteresis loops are very thin, causing small distortion. Therefore, core- type three-phase transformers behave much like three A-connected single- phase transformers as regards triple- frequency harmonics. Shell-type Three-phase Trans- formers. These could be made by FIG. 177. placing three single-phase shell- type transformers one on the other. In such a case, the leg with the coil has a width, a, while the other legs have widths a/2 (Fig. 177). FIG. 176. ! ! 236 ELECTRICAL ENGINEERING The combined intermediate sections or widths, 6, could be re- duced to n~ a, since the fluxes differ by 30 in time-phase, in adjacent intermediate sections. This is seen to be the case by noting the dotted lines in the figure. Arrows indicate what may be called the positive direction of the flux and these directions are opposite in the adjacent intermediate sections. If now the middle coil is reversed, the positive direction of the flux in the middle transformer is reversed, and the flux phases in intermediate adjacent sections have a 60 relation. The total flux in these sections is therefore exactly the same as that in the out- side section, and required width of section b is also that of the outside section namely, a/2. Transformers are therefore designed as in Fig. 178, with all width dimensions a/2, except the middle legs which have the width, a. To prove that in reversing the middle coil the flux produced is in amount the same flux as when the coil is not reversed. The flux produced by coil 1 is $ sin 0. Flux produced by coil 2, reversed, is - $ sin (0 + 120). The flux due to the two coils is then $ [- sin - sin (B + 120)] = 3> [- sin - sin 6 cos 120 - cos 6 sin 120] - $ [0.5 sin + 0.866 cos 0] = - & sin (0 + a) where a. = 30. Problem 82. Discuss the wave shapes of three-phase transformers. Show that in the core type, it makes very little difference whether the neutral is connected or not. Show that, in the shell type, the waves are essentially the same as those of three single-phase transformers. Open Delta Transformer Connection. If one of three delta- connected transformers is disabled it is possible to operate at reduced output with the remaining two, connected as shown in Fig. 179. The following is a comparison of the use of two transformers FIG. 178. MULTIPHASE TRANSFORMER SYSTEMS 237 and three transformers when the power delivered is assumed equal in the two cases. A < Current in transformer /, 7= / V Voltage across transformer. . . E p , E E TJIT Rating of each transformer. . . El v 3 ETT Rating of installation 3 /= = \/3EI 2EI Ratio of transformer capacity = -~ in favor of the three transformers. However, it may be cheaper, in a given initial installation, to buy two large transformers r than three small ones. i Now, assume a three-transformer in- / stallation in which one transformer has \ been disabled. How much should the load be reduced to give normal operation - -- of the remaining two on open delta? The line current must evidently be reduced in the ratio -p v 3 since the line current and the phase current are now the same. The output, which was -\/3EI, therefore becomes -\/3E 7= = El. V 3 ETT Therefore the ratio of outputs is x - = 0.58 or less than the V3-M ___ ratio of transformer capacity which is 2/3 or 0.0SB. b Two transformers are frequently used both for three-phase and for Teaser PLJr a combination of three-phase-two- phase transformation being con- Mam primary Mam secondary nected in a manner known as the FIG. 180. T- or SCOTT connection. T-connection of Transformers. This connection is illustrated in Fig. 180. The connection is used commonly in circuits with rotary converters, where a wire may be brought out from the neutral, h', and connected to the middle wire of a three-wire system on the direct-current side. In this case the direct current flowing in the transformer wind- 238 ELECTRICAL ENGINEERING ings has no magnetizing effect since it flows in opposite direc- tion in the two halves of the transformer windings. It consists of a so-called "main" transformer with a tap brought out at the middle points of its windings and a " teaser" transformer of 0.866 times as many turns, one terminal of which is connected to the tap, d, of the main transformer. The three- phase lines are brought to the terminals, a, b, c, which are at the three vertices of an equilateral triangle. Thus, if the base, ac, of the triangle has the length, Z, its height, bd, will be 0.866. The center of this triangle will be at a point, h, called the neutral. Rating of T-connected Transformers. Three-phase output = -\/3EI, where E and / are line voltage and current, respect- ively. Rated output of the two transformers = El + 0.866#7 = 1.866#7. .". Ratio of the output to the transformer 1 73 rating is -T = 0.925 & 92.5 per cent. This means that for the same values of E and 7, FIG. 181. three single transformers would need to have only 92.5 per cent, of the kva. rating which the T-connected transformers would have. Thus, the T-con- nection is nearly as good. It may in some cases be cheaper, as it involves only two transformers. Two-phase Three-phase Transformation. Let two-phase currents be led to the primaries, while three-phases are taken from the secondaries. Con- sidered as 1:1 ratio of the main transformers for convenience // only, the teasers would be in the' ratio 1 : 0.866. Neglecting excit- ing current the two-phase input = 2EI, = three-phase output = -7= 7 = 1.167. FIG. 182. The rating of a transformer may be taken as the average of the input and output, and it is therefore, Rating = % [2EI + 1.1QEI + (0.866# X 1.167)] = M [2#7 + 1.1QEI + El]. = El + 1.08 El = 2.08 EL MULTIPHASE TRANSFORMER SYSTEMS 239 2 The so-called cost efficiency is therefore ^-7^ = 0.96, that is. ^.Uo the rating is 96 per cent, of that of two transformers for an ordinary two-phase transformation, or for two single-phase trans- formers. That is nearly as good as using three transformers for the three-phase and, there being only two transformers, possible economy is suggested. The question arises as to how it is that, with such connections the magnetization is uniform. If it is not uniform, there will be complications due to over and under saturation in the different parts of the cores. Therefore, the sum of the magnetomotive forces due to the load current in the branches of the windings must add up to zero, that is, the two-phase load ampere-turns in each branch must be balanced by the correspond- ing three-phase ampere-turns. Let oa, ob, oc (Fig. 183), represent the three-phase AT, in amount and direction. Let de represent the two-phase AT 7 in the main transformer. To obtain the three-phase projections on the two-phase line, it is necessary to take the j? IG 183 components of oc and ob on the horizontal. These equal de. The components, however, are in opposite directions, but due to the fact that the ampere-turns from c to 6 are evidently all in the same direction, ob must be projected backward to ob'. This causes the vertical components, b'd and dc to be in opposition and they therefore cancel each other. The proof of this by trigonometrical relations is as follows: m.m.f. of primary = It sin 0, where t is the number of pri- mary turns. m.m.f. of od = ~- sin (0 + 210) m.m.f. of oe = - sin (0 - 30). Total m.m.f. = It sin + 1.16 ~ [sin (0 + 210) + sin (0 - 30)] = It sin - 0.587* [sin (0 + 30) + sin (0 - 30)] = It sin - 0.587* [sin cos 30 + cos sin 30 + sin cos 30 - cos sin 30] = It sin - 0.587* (1.73 sin 0) = 0. This means that the load current does not increase the magneti- zation of the transformer. 240 ELECTRICAL ENGINEERING In the case of the teaser transformer, both the primary and the secondary are in the same direction in space and time, that is, they bear the same phase relation as with single-phase trans- formers. Therefore, the ampere-turns relation is; secondary A.T. = 0.866 X 1.16EI = El = primary ampere-turns. There- fore as turns are proportional to voltage, the m.m.fs. are equal. It is always possible to buy transformers of both 10:1 and 9:1 ratios from stock. For practical reasons 9:1 is used instead of 8.6: 1. With these ratios, connection can be made to nearly any system in practical operation. Auto-transformer's (also called compensators). Auto- trans- formers are transformers with only one winding. The primary voltage is applied to the coil terminals; the sec- ondary voltage is obtained by connecting to taps at any desired places of the winding. The general connections of the single-phase auto-transformers are as in Fig. 184. Let /i and J 2 be the primary and secondary currents, respec- tively. Then /i = current in ab. /2 /i = current in be. The rating of the section ab is h(Ei E z ). The rating of section be is (7 2 Ii)E 2 . The rating of the auto- transformer is the average of the sum, or rating = ^ [/^ - IE 2 + I 2 E 2 - hEz] = 1 A [IiEi + / 2 #2 - 2/i^J. (Ill) Neglecting exciting current, as in any transformer, the volt- "F I age and current ratios are -^r = 7^ or IiEi = I 2 E 2 . Substi- &2 1\ tuting for I 2 E 2 in (111), the rating becomes, rating = % [2I 1 E 1 - 2/ 1 # 2 [ = I I [Ei - E t ]. The per cent, rating for a given current is Ii(Ei - Ei) = E l - Ei IjEi EI Thus, if E 2 = 90 per cent, of E ly Per cent, rating = - = 0.1, or 10 per cent. That is, it is MULTIPHASE TRANSFORMER SYSTEMS 241 necessary to supply only 10 per cent, of the rating of an ordinary transformer to effect this transformation which is obviously a great gain in cost efficiency. If the voltage is to be reduced in the ratio 2:1 the economy of using an auto-transformer instead of an ordinary transformer is not so great. The saving is in this case about one-half. Problem 83. Show the advantage of using auto-transformers by plotting a curve between per cent, rating of the auto-transformers and transformation ratio. Compensators for Two-phase Three-phase Transformation. In Fig. 185, let the two-phase taps be cb and ef, and let the three-phase taps be a, d, g, and let Ez and /2 be two-phase volt- age and current respectively and E s and 7 3 be corresponding values for three-phase. Neglecting g losses, \/3^3^3 = 2# 2 7 2 , is the power relation be- / \*\ tween input and output. ,// Considering separate parts of the windings, cur- -^ rent in ab 7 3 ; voltage in ab = 0.866# 3 E 2 , FIG. 185. since voltage in ac = 0.866# 3 ; rating of ab = (0.866# 3 - # 2 )7 3 ; current in bh = 7 2 - 7 3 . In this case 7 2 > 7 3 , E 3 being > E 2 . Voltage in bh = E 2 - H 0.866 E 3 since he = J^ac; rating of &ft = (7 2 - 7 3 )(# 2 - M 0.866^ 3 ); current in he = 1 2 7s, since the resultant sum of two equal currents 120 apart is numerically equal to one of them. The three-phase current in he is the sum of the currents of the phases hd and hg, indicated by dotted lines. Voltage in he = l /i 0.866 E 3 ; rating of he = % 0.866 # 3 (7 2 - /) ; current in de = 7 3 = current infg; voltage of de = H (#3 - #2) = voltage of fg; rating of de = or (#3 - #2) = rating of /gr; current in ec = V(/2 - h cos 30) 2 + (/ 8 sin 30") = current in c/, that is, it is 7 2 - the component of 7 3 , in phase with 7 2 + j X the component of 7s normal to Iz> p Voltage of ec = -~ = voltage of cf. Rating of ec = yV(/2 - I* cos 30) 2 + (/, sin 30) 2 = rating of cf. 16 242 ELECTRICAL ENGINEERING The combined rating, which is one-half the sum of the ratings of all the parts, is 0.933# 3 /3 ~ 0.5# 2 X V/2 2 - 1.73/ 2 /3 + / 3 2 . An examination of this rather complicated expression will show that the same ratio of cost efficiency holds with reference to the T-connected transformers having primary and secondary wind- ings, as holds for single-phase auto-transformers compared with ordinary single-phase transformers. Dissimilar Transformers in Series. Transformers may not be indiscriminately connected in series with safety. To connect two transformers of different design but proper rated voltages in series is not always safe, since they may not take their proper share of the total voltage. One may even burn out at no-load due to excessive core loss. Suppose that their normal exciting currents are different Since the same amount of current must flow through each transformer (as they are in series), this current will be insufficient to give the proper flux in one of the transformers and will be more than necessary in the other. Thus the voltages will not divide according to the rating and the core loss will be low in one and excessive in the other. Let the open circuit or exciting impedance of A, Fig. 186, be r + jx = z, that of , A ~\- JXi = Z\. B The total impedance is then Z=z + Zi = .R-f jX. FIG. 186. Then the exciting current of the two transformers in series is , _ CQ _ impressed volts " Z = R+jX The voltage across A, is Voltage across B is V B = e Neglecting the power component, we get as a fair approximation, MULTIPHASE TRANSFORMER SYSTEMS 243 snce ' approximately, A - _ /.I I m ,A where I m ,B> and 7 mrA , are the normal magnetizing currents. Thus, the respective voltage drops across A and B, when in series, will be approximately inversely proportional to the normal magnetiz- ing or exciting currents. This assumes constant values of x and rti, which would not be true if the resultant exciting current differed widely from the normal values of exciting current of the two transformers. If x is nearly equal to x\ 9 the above ratios would hold. Where one transformer, however, is saturated, its reactance is greatly diminished, which allows a greater current to flow in the circuit but tends to equalize the voltages. Dissimilar Transformers in Parallel. This is the usual mode of connection and it offers no difficulty due to unequal exciting currents. The question, here, is one of proper division of the load. In giving orders for additional equipment, it is customary to specify what the percentage reactance of the new transformers shall be. With equal, or proportional, reactances there results a proper division of the load. Consider the parallel connection as shown in Fig. 187. The two load currents are: I A = i + Ji', IB = i\ + fi'i- The total load current, I = I A + I B . ' ' Let Z A = r + jx and Z B = TI + FIG. 187. be the impedances of A and B respectively. Then, since the terminal voltages are the same on each, the voltage drops in the transformers are equal, and are: (i + ji')(r + jx) = (ii + ji'i)(ri + jxi). 244 ELECTRICAL ENGINEERING Multiplying out, ir + jix + ji'r - i'x = itfi + ji&i + ji'tfi - i'&i. Here, the real components must be equal and the imaginary components must be equal. .". ir - i'x = itfi - i'&i, and ix + i'r = iiXi -j- zVi. Neglecting resistances, and ix = whence / i> X\ i> 3/1 z'i "~ a; ' ii "~ # Thus, the load is divided in inverse proportion to the reactances. The best method of connection is, as in Fig. 188. The student is advised to explain why this is so. ' A ^c ft P i FIG. 188. Three-phase Connection of Dissimilar Transformers. If the three transformers are connected Y Y there will not be symmetrical distribution of voltage. Consider the neutral point, 0, Fig. 189, with reference to the transformers A and B. With line voltage impressed on AB, the potential at may have any intermediate value, just as with two single-phase trans- formers in series, depending on the relative open circuit im- pedances of the two transformers. The point of junction of the three transformers may, for in- stance, be displaced to 0'. The secondary Y-voltages would have a similar relationship to each other. Dissimilarity may consist merely in variation in the iron of two supposedly similar transformers. If the secondaries are connected in A, the primaries being Y-connected this difficulty of unbalanced potentials is eliminated. The induced voltage in each secondary will, of course, be MULTIPHASE TRANSFORMER SYSTEMS 245 proportional to that of its primary, giving the closed A, ABC, Fig. 190, when the primary circuit is balanced. With unbalanced condition, if the A is left open at B, the vol- tage vectors will not make a closed figure, but as shown by the dotted lines, will leave an opening between B' and B". If the A is then closed, the voltage B'B" will act in the A circuit, sending a local current which will increase the magnetization of the transformer whose flux is below normal and decrease that of the transformer whose flux is above B' FIG. 190. FIG. 191. normal. Thus the magnetization is brought back to normal value, the local current in the A serving to anchor the neutral point of the Y. Three-phase transformer systems may be extended in a variety of ways to cover cases where it is desirable to use six phases. This practice finds application especially with rotary, or syn- chronous, converters, and it will be discussed more fully under that heading. Such combinations of transformers as permit symmetrical grouping of voltages are illustrated by the double A, double T, or double Y shown in Fig. 191. i 2 3 (a) CHAPTER XXXIII ALTERNATORS Fundamentally, direct-current and alternating-current genera- tors are alike. An alternator becomes a direct-current generator by adding a commutator. The essential principles of both machines have been developed in Chaps. VI and VII. In Fig. 192, a, is represented a simple alternator with a two-pole field core magnetized with direct current from some independent source, and an armature with a single coil. As this armature revolves there is generated in it an e.m.f. which follows closely a sine wave of time values. It is apparent that the space on the armature periphery is not all utilized, and that another coil could be put on in space quadrature to the first. In such a case, two similar e.m.f. waves would bfi produced but in time quadrature with each other, or at 90 time-phase displace- ment, that is, one wave would reach its maximum one-quarter of a period later than the other (Fig. 192, 6). Such an alternator is called a two-phase, or, sometimes, a quarter-phase machine. On the same principle, an armature may be supplied with three coils, or groups of coils, spaced 120 apart, each group giving its separate e.m.f. wave (Fig. 192, c). In this way, any number of coils or groups of coils may be wound on an armature, giving any desired number of phases. In practice, however, the majority of alternators are three- phase, and very seldom is one built for a greater number of phases than three. The voltages generated in the various phase windings may be conveniently shown in their proper relations by vectors. If in the two-phase case, the ends 1' and 2' (Fig. 193, 6), are 246 (6) FIG. 192. ALTERNATORS 247 joined together, the voltages of the two coils will be added vectorially, so that a voltmeter placed across the terminals, 1, 2, would read \/2 times the voltage of either coil taken separately, since the two voltages are in time quadrature. Likewise by connecting 1 and 2', the joint reading across 1' and 2 will obvi- ously also be \/2 times the voltage of one phase. With a three-phase machine it is not quite so apparent that the voltage between the three collector rings is \/3 times the voltage generated in one phase. At first sight it might be expected that the resultant voltage should be the same as that generated in each phase since the voltages are 120 apart. C FIG. 194. Let, in Fig. 194, OA, the voltage of phase A, be represented by e A = E m sin at. Then OB, the voltage of phase B, is evidently e B = E m sin (at + 120), and e c = # m sin (orf + 240). The voltage between collector rings A and B is thus SA e B = E m [sin at sin (co + 120)], which, by simple trigonometric transformation becomes, A _ B = V3#m sin (w< - 30). Thus the numerical value of the potential difference between the collector rings is \/3 times as great as the voltage generated in each phase and the resultant voltage is displaced 30 from the voltage generated in phase OA. Problem 84. Prove that the voltage, between B and C is: \/3E m sin (wt + 90), and that the voltage between C and A is: \/3E m sin (ut + 210). It is seen, thus, that the voltages between the collector rings are also 120 apart. 248 ELECTRICAL ENGINEERING It is interesting to note here that a single-phase machine might be treated as a two-phase machine in which the two phases are 180 apart as is shown in Fig. 195. Let the voltage generated in OA be E m .sin co. Then that ^ generated in OB is E m sin (coZ + 180). Thus ~B~~ o ~~5 the difference of potential between the collector FIG. 195. rings at A and B is (sin (at sin (' + 180) = 2E m sin ut. The resultant potential difference is twice the voltage generated in each phase, as should, of course, be the case. This fact could have been developed also geometrically. To find the potential difference between A and B } Fig. 194, we should subtract OB from OA as shown in Fig. 196. It is not necessary that the windings shall consist of separate coils. A closed ring winding, or Gramme ring, may be tapped at symmetrical points and these connected to slip rings, as in Fig. 197. Thus, if a voltmeter is connected across the slip rings (1,1), the voltage of one phase is read. If connected across rings (2, 2), the same value of voltage will be indicated, but it is evident that the phase of this e.m.f. is displaced by 90 time degrees from that of the first. FIG. 196. FIG. 197. With the three-phase connection taps are brought out at points 120 space degrees apart and led to slip rings. A volt- meter, connected across any two rings, will read the voltage of one coil, say coil a, Fig. 197. But this must also be the sum of the voltages of the other two coils, since any one coil is in parallel with the other two coils with respect to the external circuit. The voltages in this case form a closed, so-called delta, A, and it is evident that the phase voltage and line voltage, or voltage between the collector rings, are equal. In these diagrams only three collector rings and three lines are shown. Yet in the discussion it has been assumed that one side ALTERNATORS 249 of each winding is connected to a common point. It would seem, therefore, that at least four collector rings and lines might be necessary to form a complete system, in other words, that even a balanced three-phase system would involve four wires as is shown in Fig. 199. It is evident that if, with a balanced system, the current in an ammeter placed at N is always zero, then no return or fourth wire is necessary. FIG. 198. FIG. 199. Let the current in phase A be ia = Im (sin a>t) and the current in phases B and C be i b = I m sin (orf + 120) and i c = I m sin (at + 240). The current in N is then in = ia + ib + ic = Im sin 0=0. Problem 86. Prove that no circulatory current of fundamental frequency flows in the delta-connected generator. Since with the Y-connected generator the transmission lines really form extensions of the windings, it is evident that whatever current flows in the line also flows in each winding. With the delta-connected generator this is not so, because the line current is the vector sum of the currents in the adjacent phases, as is shown in Fig. 198. The current in phase 1-2 may be considered the zero vector. Thus the currents in the phases are: in = Im sin (at, i* = I m sin (at + 120), iai = Im sin (tat + 240). Then, since the sum of the currents flowing to a point is zero, it + las - in = 0, or it = iiz iza I m sin wt I m sin (wt + 120) = Vzlm sin (cat - 30). The line current is thus \/3 times as large as the current in the individual phases. Referring to Fig. 198, it is evident that is + in *32 = and i\ + in isi = 0. Problem 86. Prove that the currents in lines 1 and 3 are, respectively, sin (tat + 90) and \/3/m sin (tat + 210). The power given by a three-phase alternator is P = \/3EI cos a, 250 ELECTRICAL ENGINEERING whether the alternator is connected Y or A, where / is the effec- tive value of the line current and E the effective value of the voltage between the lines, and a is the angle of lead or lag of the phase current in reference to the phase voltage, that is, cos a is the power factor. To prove this, consider a Y-connected generator. Since I is the line current, it is also the current in each winding. Since E is the line voltage, the voltage of each of the three E phases of the generator is =? Thus the power given by each of V3 pr the three phases of the generator is = cos a, and the total V3 El power, 3 -= cos a, = \/?>EI cos a. v 3 Problem 87. Prove that this also applies in the case of a delta-connected generator. Voltage to Neutral. In Y-connected alternators the neutral point is the center of the Y. On a three-phase distribution system it is often advantageous to run a fourth wire from the neutral. The voltage between any of the other wires and the neutral is the phase voltage, and is equal to the line voltage divided by Vs. In a A-connected alternator there is no actual neutral point. However, for purposes of calculation, a neutral point is imagined at the center of the delta, and the voltage to neutral is then the phase voltage divided by \/3, or, since the phase voltage and the line voltage are the same, it is equal to the line voltage divided by \/3 as with Y-connected alternator. The voltage to neutral is thus independent of the manner of connecting the alternator windings. Rating of Alternators. As with transformers, alternators are rated in kilovolt-amperes, not in kilowatts. This is because the permissible output of an alternator depends on the current in its windings, regardless of the phase relation between the current and the voltage. The nominal rating of an alternator may be designated as, for example, A.T.B. 12-400-600-2300, where A signifies alternator, T signifies three-phase, (S is for one or single-phase), B signifies a ALTERNATORS 251 revolving field. If the armature is the revolving part, the third letter is omitted. 12 signifies the number of poles. 400 signifies the rating in k.v.a. 600 signifies the speed in r.p.m. 2300 signifies the rated voltage. Sometimes a subscript is added to the second letter; thus, A TZ signifies a three-phase revolving armature alternator having two slots per pole per phase on the armature. If the above alternator is Y-connected, the phase voltage, or 2300 voltage to neutral, is -1= 1330 volts. 400 k.v.a. 400,000 The line and phase current is 23QQ = 3^1330 = 10 amp. If delta-connected, the voltage to the imaginary neutral is likewise 1330. 100 The line current is also 100, but the phase current is /= = v 3 57.7 amp. CHAPTER XXXIV ARMATURE REACTION The so-called armature reaction of a machine is a measure of the m.m.f. of the armature. It is thus expressed in so many ampere- turns, either on the whole circumference of the armature or, more often, the m.m.f. on one pole of the armature. This latter convention will be used in this book. As will be seen, the m.m.f. of the armature current sometimes acts against the m.m.f. of the field excitation, sometimes it assists it, and often its effect is only to shift the flux. In Fig. 201 the coil (1, 1) is in the position of zero, or mini- mum, e.m.f., assuming the flux to be symmetrical in the field system, or due to the field ampere-turns alone. The coil (2, 2) is in the position of maximum e.m.f. This condition may be assumed to hold for no-load. The coil (3., 3) is in an interme- diate position. The current may or may not be in time-phase FIG. 201. FIG. 202. with the e.m.f., but whatever its time-phase relation may be, in spa6e, it is evident from Fig. 201 that the m.m.f. of the coil is at right angles to the surface of the coil, and therefore at right angles to the line which represents the position of the coil. In Fig. 202 let the armature current lag behind the e.m.f. Its m.m.f. is seen to be largely in opposition to that of the field which causes the main flux, and this opposition increases the greater the lag and becomes complete at 90 lag. Similarly a leading current 252 ARMATURE REACTION 253 umru S" FIG. 203. increases the flux. The current, i, may be divided into two. components, one of which, i", is entirely wattless and exactly opposes the field flux, and the other, i 1 ', the watt component in phase with e { , which merely distorts the field. The effect of current in the armature is to weaken the resultant flux and to displace its maximum position, if the current lags, as shown in Fig. 203. The weakening is due to the wattless component, the displacement or distortion to the power component. Thus the trailing pole-tip may even become saturated, while the leading pole-tip is robbed of a large part of its flux. The position of the coil for maximum induced e.m.f. is shifted ahead, with lagging current, and behind, with leading current. These relationships are shown in Fig. 204, where the induced e.m.f., e iy is taken as the zero vector. 6i is at right angles to the resultant flux, fa, and lags behind it. The armature current, /, is taken at any angle and produces a flux (f> a in time-phase with it. This flux vectorially subtracted from fa, gives f and fa, or, 7 = /? 90, where is the angle between e t and i = - i where pi = -r- = reluctance of a Pi dx 270 DETERMINATION OF THE SELF-INDUCTION 271 small path of length 6, in air, and of cross-section, dx sq. cm. (dx X 1). The reluctance of the iron is neglected. Then, x dx - -=- a o The interlinkages or turns linking with this flux are n- Hence, the flux-turns interlinkage per unit current, or the inductance across the width, a, of the coil, is 1 C a n 2 x 2 a n 2 Li = j I 4ir-^I -^dx = g47ry, per cm. length of effective iron. 2. Inductance Due to the Flux through Section c. The mag- netomotive force is F 2 = nl. p 2 = The flux is c P2 All the n conductors link with this flux. 3. Inductance Due to the Flux across the Section d. This, by a similar process, is e 4. Inductance over the Face of the Tooth. The magneto- motive force is F 4 = nl. The reluctance, p\ = -r- The flux set up is All the n conductors link with this flux. 5. Inductance of End-connections. The inductance of the part of the winding that projects outside of the iron is almost impossible to estimate accurately. It depends largely on the mechanical design. If the end shields are some distance away from the winding, a fair approximation is obtained by assuming the flux per ampere-turn per centimeter of wire to be inversely proportional to the square root of the pole pitch, or what is equivalent, the square root of the armature diameter divided by 272 ELECTRICAL ENGINEERING the number of poles. A fair approximation to the flux per ab- solute ampere-conductor, per centimeter of wire is ' 5 = 13\JY) ( Y) (or 1.3 A when amperes are used), where p is the number of poles, and D = armature diameter. If the length of the end-connections of a coil, counting both sides of the core, is 8 X > then the flux per turn, per ampere- conductor, is Thus, with a single-coil winding, where all conductors of a coil are wound together, the flux per coil is 05 = 104n/ X \ > Mp where n is the number of effective conductors and / is the current per conductor in absolute amperes. The inductance, L 5 , will be due to only one-half of this flux since each coil occupies two slots. .-. L b = ^- = 52n< I D A/ \ p The total inductance for a single slot exclusive of end-connec- tions is then, per centimeter net length of armature iron, and the total inductance, including end-connections for length, I, of iron, is, in henrys, a , c , d , / 13 ID 10 9 ,__. For a three-phase alternator, with one slot per pole per phase, there is also to be added a term due to the flux, 6 , in parallel with 04, which passes from the next adjacent half tooth, across the gap (Fig. 213). FIG. 213. The inductance due to this flux will vary greatly, according to the air gap, whose cross-section and length may be very different from the values used in determining L 4 . DETERMINATION OF THE SELF-INDUCTION 273 Problem 93. Calculate the reactance per phase of the following alter- nator when the slots are under the poles. A.T.B. 8-100-900-2300 v. 48 slots; 24-in. armature diameter; therefore, 2 slots per pole and phase ; 28 effective conductors per slot. Dimensions, referring to Fig. 214, are: a = 1 . g = average gap under adjacent teeth = 0.25 6 = 0.75 g' = average gap under distant teeth = 0.5 d = 0.14 n = 28 e = 0.27 s = 2 / = 0.10 p = 8 h = 0.85 I = 9 k = 0.82 D = 24 The wires are confined to the distance, a, of the slot. Each slot has n effective conductors and there are s slots per pole per phase. Let 0i be the field which crosses the conductors due to the m.m.f. of the coil which is between the bottom of the slot and the distance X. Then the m.m.f. is *,/, where 7 is the current in amperes in the conductor. The flux, in section dx per cm. depth of magnetic circuit parallel to the shaft, is therefore, 4?rFi _ birsnxl Pi Pi But the reluctance pi of the path is Pi = -r- neglecting the iron. 18 274 ELECTRICAL ENGINEERING Thus, 4wsnxldx 4:wnlxdx dd>i = r = T * asb ab M This flux interlinks with - cs conductors. Therefore the inductance of this part of the magnetic circuit, which is the interlinkages of the turns and flux across the conductors per unit current, is ^a 4.Trn 2 s T 7 a Ix 2 dx = 47rn 2 s ^r- 3o Consider next the inductance of the part of the magnetic circuit which is above the coil proper. The m.m.f. is that of all conductors, and is F z = snl. The flux 02, per cm. length, is Pz P2 In this particular case there are three magnetic paths in multi- ple, the first, of reluctance, -j> the second, of reluctance, > and (t 6 sb the third, of reluctance, y 1 ;; 1 . 1 1 _d+f e_ * " 2 sb sh sb sb sh and d +f Some flux crosses the two gaps from the teeth adjacent to the coils and causes an inductance which is similarly determined. Thus, the m.m.f. is F 3 = snl. 4irsnl k PS 2g 7 1 A T k . . L 3 = j 4TrsnI pr- sn L zg DETERMINATION OF THE SELF-INDUCTION 275 Similarly, the flux which crosses the gaps from the more dis- tant teeth causes an inductance, L 4 = 47Tsn 2 -r Thus, the total inductance per centimeter depth of magnetic circuit covered by the iron is T t 9 F a , d -f / , e , ks , ks~\ L = 4?rm2 L5 + V + 1 + *, + w\' If I is the net length in centimeters of the iron of the armature core, and p is the number of poles, then the inductance per phase of the part of the electric circuit which is in the slots is , 7 r a d + / . e . ks . ks~\ Lo = 4r.pl [36 + V + h + 2g + W\' (The dimension of inductance and capacity in the absolute system of units is centimeters.) By extending the reasoning in the case of a single slot, the in- ductance of the end-connections per phase is found to be L 5 = 52sWp\ = 52s 2 ra 2 \ p With bar winding, when the coils are split up, as shown in Fig. 215, the inductance of the end-connections becomes L 5 = 77 9 / 13s 2 n 2 \/Dp, since the m.m.f. per end coil is ^-t and the inter- linkages are ^- The inductance of the ma- / \ chine per phase is then L = Lo + L 6 in cm., or r LO ~r LS , L = ^ henrys. If inch measurements are used, FIG. 215. f . e . ks and L 5 = 83s 2 n 2 \/Dp for single coil winding, or = 20.8s 2 n 2 \/T)p for split coils, and the inductance in henrys is L = 276 ELECTRICAL ENGINEERING Applying these equations to the particular three-phase alter- nators given above, and noting that the coils are not split up, we get: Lo = 32 X 2 X 28 2 X 8 X 0.9 [3^75 + ^ + 5^ + (0.445 + 0.32 + 0.33 + o^ 1 _ 2 + q : ^x_ 2] = 21)600j000cinj 3.27 + 1.64) and and L 5 = 83 X 4 X 28 2 X Vl92 = 3,600,000 cm. .'. L = 25 > 20 ^ 000 henrys = 0.0252 henrys. x = 27r60 X 0.0252 = 9.5 ohms. This is, then, the reactance for the slot under the pole, that is, the reactance which should be used with the power component of the current. The reactance be- tween the poles is less and may be taken as 0.6z or xi = 5.8 ohms. It is very convenient in designing a slot, to make it accommodate four coils. As this is a very common arrangement, the calculation of the inductance of a single tooth armature having four coils in the slot is also made. The cross-section of the slot is shown with dimensions in Fig. 216. The procedure is practically the same as in the preceding case. The flux through a small section, dx, of the space occupied by the lower pair of coils is x N where the magnetomotive force is F x = -^I,N being the total (i & number of turns. Thus, the flux-turns interlinkage per unit current or the induct- ance through the lower pair of coils, LI, is dx ba DETERMINATION OF THE SELF-INDUCTION 277 The inductance across the insulation, h, between the layers is ik? L2 = ^_4 _ h b b The inductance across the upper coils is 1 f 4 'T J1+ g /JL -r a b a _ SrrJV^a ^ TrN 2 a _ 7irN 2 a o ~ 36 36 36 The inductance across the insulation, c, beneath the wedge, is The inductance across the wedge is L 5 = The inductance across the face of the tooth is The inductance of end-connections is, as in the previous case, L 7 = 527V 2 J \ p The total inductance per centimeter effective length of core is Problem 94. A certain three-phase, 60-cycle alternator has one slot per pole per phase. The dimensions in inches of slot, etc., are as' in Fig. 216, where a = 0.45, 6 = 0.75, c = 0.14, d = 0.37, e = 0.85, / = 0.82, g = 0.15, h = 0.1. There are twenty-four slots, thirty-two effective conductors, the effective length of the armature core is 9 in., the armature diameter is 32 in. Show that the armature reactance is approximately 4 ohms. CHAPTER XXXVII FIG. 217. ARMATURE REACTION IN MULTIPHASE MACHINES With current in the armature of an alternator, two magneto- motive forces exist, one, that of the field winding, and the other, that of the armature winding. Sometimes these add directly but more often they are more or less in opposition. If the resultant field flux is in the direction of the field poles, Fig. 217, and the armature winding is assumed concentrated in a coil in position a-b, then the induced e.m.f. due to the rotation of the coil in the field is e { E m sin and the current is i = I m sin (6 + a), where a is the angle of lead of the current in respect to the e.m.f., that is, tan a = -, where x and r are the total reactance and re- sistance of the external and armature circuits, and EM and IM the maximum values of the e.m.f. and current respectively. Jf the armature coil has T turns, the m.m.f. of the armature is obviously, iT = JJTsin (B + a), In the position shown the m.m.f. of the armature does not act in line with the m.m.f. of the field winding, but its component in the direction of the field is a' - &' or the total m.m.f. multiplied by cos 0. The component b - b' of the armature m.m.f. at right angles to the field is, of course, the total m.m.f. multiplied by sin B But this component does not increase or decrease the field, but only distorts it. 278 ARMATURE REACTION 279 Let M be the component of the armature m.m.f. in the direc- tion of the field m.m.f. Then M = I m T sin (0 + a) cos = I m T (sin B cos a + cos sin a) cos = I m T(% sin 20 cos a + cos 2 sin a). But COS2 e = L-*. /mT 7 ' M = ~T~ t sin 2e cos a + sin a cos 2 + sin 1 7 T = -y- [sin (20 + a) + sin a]. It is seen that the average value of the armature reaction in the T rp direction of the poles has a constant value which is -^- sin a, 2i and superimposed upon this is a pulsating reaction, a m.m.f. which pulsates at double frequency. The effect of the latter is zero when considering the average effect over a cycle. T rp IT J-mJ- . . M av . = y sm a, But I m sin a is the maximum value of the wattless component of the current (Fig. 218). Thus the armature m.m.f. (or armature reac- tion, as it is called), in the direction of the poles corresponds to the wattless component of the FIG. 218. current. Thus, if the current is in time-phase with the induced e.m.f. (in which case there is no wattless component), the armature current neither magnetizes nor demagnetizes the field, but only distorts the distribution of the flux. If the armature current leads the induced e.m.f., then it is seen that the armature reaction is positive. It helps the field m.m.f. If the current lags, then a is negative and the armature reac- tion opposes the field m.m.f. In a three-phase machine the e.m.fs. of the different phases may be expressed as ei E m sin e 2 = E m sin (0 + 120) e 3 = E m sin (0 + 240). 280 ELECTRICAL ENGINEERING Prove that the average armature reaction in the direction of the poles is 1.57 m T sin a, and is not pulsating but steady. NOTE. In specifications of alternators one item is usually called armature reaction and the value given is ~ o~> in a single-phase machine, I m T in a two-phase machine, and l.5I m T in a three-phase machine. In this case, however, I m is the maximum value of the rated current, and T is the effective number of turns per armature pole per phase. Example. Find the so-called armature reaction in an 8-pole, 100-kw., 2300- volt, three-phase generator which has 224 armature turns per phase and which is Y-connected. Answer. The voltage per phase is = 1330. The full-load effective current is 100,000 ^=- - = 25.1 amp. \/3 X 2300 .'. I m = 25.1 \/2 = 35.5 amp. The winding is practically concentrated so that all turns are effective, thus .'. M a = 1.5 X 35.5 X 28 = 1490 A.T., and this is the numerical value given to " armature reaction." If the armature actually carried full-load current and the cur- rent was lagging 90 time degrees behind the e.m.f., and hence was 90 space degrees displaced from the main field flux then the de- magnetizing ampere-turns would be 1490. If the current was leading then the armature current would assist the field to the extent of 1490 A.T. With a phase angle, say 30, the actual magnetizing or demag- netizing ampere-turns would obviously be only 745. In an n-phase machine the armature reaction is not pulsating but has a constant value, M Im M a = - ^ - Sm a ' Consider any particular phase indexed m. Its voltage is e = E m sin (0 + - J ; ARMATURE REACTION 281 its current is ; = 7 m its m.m.f. is M = iT = 7 ra T sin cos The total m.m.f. at any instant is, thus But, m = n M = S m = 1 in (20 + sin + = sin (26 + a) cos cos (20 + a) sin The sum of all terms containing cos must n be zero, because sides in a closed the sum of the cosines of all polygon is zero. Similarly the .. are zero. Thus it fol- sn n terms containing lows that, M a = M (since it is constant for all values of 0) = n FIG. 219. sin a. FIG. 220. Effect of Distributed Winding on the Armature Reaction. Consider a single- phase armature wound with a number of coils as is shown in Fig. 220, 6, all of whose coils are connected in series. The effective value of the e.m.f. gener- ated in coil A may be represented by.OA. The e.m.f. in coil B is then represented by AB, and so forth. It is seen that in this case the resultant e.m.f. is less than the algebraic sum of the individual e.m.fs. of the coils. It is the vector sum of the e.m.fs and is 2/7r times the algebraic sum. If the total winding has N turns, the equivalent number of turns of a concen- trated winding would be T = 2/ir N. 282 ELECTRICAL ENGINEERING If instead of being distributed all around the periphery the winding covered an arc of, say, 60, as is shown in Fig. 221, the effectiveness would again, by a similar diagram, be found to be the ratio of the chord to the arc. Thus, the chord is evidently 2 sin 30 and the arc -- /" A\. \ ,. fc _ i-taW _ 8 f g;sft _.. 7T T and T = 0.955AT FIG. 221. In general, if the winding covers a electrical degrees, a 360 ir Example. A completely distributed single-phase winding has a = 180. .'. * = 2 - 7T Three-phase winding uniformly distributed. In this case, the winding covers 60. Thus, k = 0.955. CHAPTER XXXVIII HUNTING The periodic oscillation of synchronous machinery is a familiar and oftentimes troublesome phenomenon, It manifests itself principally by the swinging of the needles of meters connected in the circuits. When the effect is cumulative, it continues to in- crease until rupture occurs somewhere in the system. Often it is not cumulative, and resembles simply the movement of any vibrating body such as a pendulum. The difficulty of visualizing hunting of a revolving machine comes from the fact that the vibration is superposed on the steady rotation of the moving part. It can be well imagined as similar to the motion in space of a pendulum swinging east and west while at the same time the earth, on which the pendulum is fixed, is in rotation. Hunting of electrical machines is possible because the position of the armature core in the field structure at any moment is deter- mined by the balance of mechanical and electromagnetic forces. Assuming the mechanical force to be steady, as represented by the shaft or belt in connection with the prime mover or load, the electromagnetic force is variable owing to the highly elastic property of the magnetic field. Under absolutely steady condi- tions there would, of course, be no hunting. But such conditions do not exist, and any variation of the electromagnetic forces results in a change of speed as the machine re-establishes the momentarily lost equilibrium. Hunting, or oscillating, is thus started and continues as equilibrium is gradually restored in the elastic medium of the field. The mechanical force is not always steady. Steam engines, and especially gas engines, are subject to pulsation of driving torque. This may appear in the generator in the form of forced electrical vibrations, especially where the machines are directly connected. When the generator is free to oscillate in response to any impulse, it does so at a definite rate called its natural period in distinction to a forced period. 283 284 ELECTRICAL ENGINEERING The natural period of a pendulum depends on its length and mass, the length being the radius of gyration. Similarly the natural period of an armature or revolving field structure depends on its mass and radius of gyration. To find the natural period of a machine, consider the motion of a stretched spring as illustrated in Fig. 222. The spring suspends a weight, and its motion is damped by a piston working in a dash pot. Let F = pulling down force in the spring, y = dis- placement of the weight. Then, f t = ay = tension on spring, where a = number of pounds, per unit length, of the downward pull. If the friction force due to the dash pot is assumed proportional to the velocity, the force necessary to overcome friction = // = k -T- (The power required varies as the square of the velocity.) The force required to overcome inertia = M or, where Weight Dash Pot FIG. 222. M = mass and a = acceleration, or, . . = M = ay M is the total force required to balance those acting in the system. If the applied force is removed, or if F = 0, the equation becomes, dt = 0. Applying this equation to an alternator, ^-^yy^r^nm^-^ the condition is as illustrated in Fig. 223. The moment equation is Pp = I _}- - _|_ a Q } FIG. 223. where Fp = the applied moment, F being the force and p the lever arm. 7 = moment of inertia, 6 = initial angular displacement, )3 = moment of retarding force per unit angular velocity, a = twisting moment per unit angular displacement. dB -J7 = angular velocity, HUNTING 285 -77 = moment of angular velocity. p = radius at which the force is applied. la = moment of angular acceleration. This is, by ordinary mechanics, Md 2 s d ds When the force is released, Fp = 0, and The solution of this differential equation is = Ae*V + Bt m J, in which A, B t mi and niz are to be determined from known conditions. Let dO Then, m 2 ! + 0m + a = 0, arid 2_L^ m z -\- j m = jp> whence, _ - ~ 21 - 2l 4P ~ I ; m2 = " 27 If vp j is positive, then 5 is real, and the equation shows that 6 gradually decreases to zero without oscillation. If, however, j3 2 a jp -j is negative then the square root is imaginary and 6 reaches zero after a certain number of oscillations. 2 a Thus, hunting can take place only when ^ j is negative. Let then Then ._.__. 5 = 7~4/ 2 /3 mi = - + 286 ELECTRICAL ENGINEERING and a m * = ~ 27 ~ jd or putting Wi = - 7 m 2 = - 7 - .'. = Ae-^e js When t = T, the time of one period, 6T = 27T Assume the case of suddenly throwing off full-load from the alternator. Then 6 = . At t = 0, the hunting has not yet begun. " The period, T=^= , M , ^ VW - |8 2 where /3 is the friction torque, and has little influence on the period of hunting, but rather affects the amplitude. We may assume = 0. Then where T 7 is in seconds. The beats, or oscillations per second, are > or oo Beats per minute = * 7T \ x The angular space position of the alternator armature with reference to the field pole may be determined for any load (Fig. 224). Let this angle be assumed to be 20 for a two-pole machine HUNTING 287 at full-load, or 10 for a four-pole machine. If 6 = mechanical angle, and = electrical angle, 8 = where v = P number of poles. Torque, 7050 X kw. . r.p.m. If a = torque per unit angular displacement, T 57.3 where = angle in degrees for the load being considered. Therefore, = 24.25 X kw. X / X 10 s where / = frequency, N = revolutions per minute and is in degrees. Finally, the solution is, Beats per minute, S m = The number of beats per minute may be changed by changing / or , the former by the addition of a fly-wheel, the latter by altering the gap. Bridges, or dampers, between the poles may also be used to produce eddy currents for the purpose of damping the oscillations. Problem 96. Determine the periods of the 100-kw. alternator of the previous problems, both as definite pole and as a round rotor machine, and with long and short gaps. Solution. The equation is = 4^000 /kw. X/ in which the constants previously given are. N = r.p.m. = 900; kw. = 100; / = 60; TPr 2 800* X 0.86 2 1 = moment of inertia, = = ~ oo i A p = 0.86 ft. = radius of gyration. 288 ELECTRICAL ENGINEERING Supplying numerical values, 47,000 /100 X60 \18.4X d> 942 \ -o tan /3 = tan (5 900 \18.4X<^ To find 0, in Fig. 224, = - 90 + a. Assuming non-inductive load, Ei = e + Ir + jlx = a + jb. F/ = be ml -f- ,/aC = d + jf. f_ _b tan 5 tan a _ d a af bd '' 1 + tan 5 tan a = bf_ = ad + bf + ad where a = e + Ir = 1330 + 25 X 0.69 = 1347.25, d = -bC - ml = -25zC- 25m, / = a <7 = 1347.25C. Tabulating, for the four cases : Definite pole Round rotor Gap, in. 0.25 0.1875 0.25 0.1875 X 8.15 14.1 8.15 14.1 C 2.75 2.18 2.75 2.18 m 47,5 47.5 59.4 59.4 f 3,700 2,940 3,700 2,940 b 204 353 204 353 d -1,750 -1,960 -2,040 -2,250 of 4,980,000 3,960,000 4,980,000 3,960,000 bd -357,000 -692,000 -416,000 -795,000 ad -2,360,000 -2,640,000 -2,745,000 -3,030,000 bf 755,000 1,040,000 755,000 1,040,000 af - bd 5,337,000 4,652,000 5,391,000 4,755,000 ad + bf -1,605,000 -1,600,000 -1,990,000 -1,990,000 tan/9 -3.32 -2.92 -2.71 -2.39 /3 106.75 109 110.25 112.7 tan a 0.1514 0.262 0.1514 0.262 a 8.6 14.67 8,6 14.67 ft - 90 + a 25.35 33.67 28.85 37.37 1/0 0.0394 0.0297 0.0346 0.0268 Vl/* 0.190 0.172 0.186 0.1635 s m 187 162 175 154 CHAPTER XXXIX STUDY OF THE DESIGN CONSTANTS OF ALTERNATORS Alternators differ primarily in respect to the number of phases, and whether the armature or the field structure is the revolving part. Secondarily, they differ in respect to the frequency, voltage, output rating and speed. In practice, the very great majority of alternators are of the three-phase, revolving-field type. In frequency, they are gen- erally of either the 25-cycle or 60-cycle type in America; 25- and 50-cycle in Europe. Voltage may be any desired value up to about 13,000. In output rating alternators are built up to 30,000 kva. The speed is limited by the prime mover and the frequency. Maximum speed, for 60-cycle machines is 3600 r.p.m., corre- sponding to the requirement of a bipolar field; for 25-cycles, the maximum speed is 1500 r.p.m. The chief types of prime mover used with alternators are the reciprocating engine, representing moderate speeds, the water turbine representing low speeds," and the steam turbine representing high speeds. Certain roughly approximate constants have been obtained from experience which may serve as guides in preliminary design. These are given in Table IX. TABLE IX. APPROXIMATE CONSTANTS OBTAINED FROM EXPERIENCE Prime mover Recip. engine Water turbine Steam turbine Frequency. . . 25 5 3,200 2.5 6 2.5 60 3 1,800 2.5 6 2.5 25 13 8,500 2.5 6 2.5 60 5 3,200 2.5 6 2.5 25 20 13,000 2.5 6 2.5 60 7.5 4,800 2.5 6 2.5 Arm. dia. per pole Arm. reac. per pole No load A.T.-per pole Arm. reac. Regulation (approx.), per cent Sh. cir. cur. at load exc. Full-load current 19 289 290 ELECTRICAL ENGINEERING Using them as a basis, the design constants will be calculated for the following alternator: A.r.B-8-100-900-2300 volt. General Constants. From the rating it is seen that the machine is a three-phase, revolving-field, 8-pole, 100-kilo volt-amp., 900- r.p.m., 2300-volt alternator, evidently to be driven by a recipro- cating engine. It is first necessary to decide whether the phase windings shall be connected Y or A. Y-connection is, in general, suitable for higher voltages and lower currents. Therefore Y-connection will be assumed in this case. The phase winding voltage is then The phase current = line current Kva 100,000 ^ r.p.m. ^, poles 900 v Frequency = ^ Q X ^ = -QQ X 4 = 60 cycles. Slot Dimensions. The development of the design now depends on the determination of size and number of slots and the conduc- tors *in the slot. It has been found that for an n-phase machine, armature reac- tion per pole = ~ I P t, where t = effective turns per pole and phase, z and Ip is the maximum value of the current in the windings. For three-phase, therefore, by Table IX, 1800 amp.-turns = 1.5\/2 X 25. . ' . t = 34. This number serves as a good preliminary value. Actually, 28 turns per pole per phase were chosen. Conductors per pole and phase are then 2 X 28 = 56. The number of slots per pole per phase depends primarily on the armature circumference and the slot pitch. With many slots, a smoother e.m.f. wave is generally obtained. The number of slots is also usually greater in low voltage machines, where the requirements of higher insulation are not so severe. Practically, at least two slots per pole per phase are used. DESIGN CONSTANTS OF ALTERNATORS 291 From Table IX, the armature diameter per pole is found to be 3 in. Hence the diameter is 3 X 8 = 24 in. and the circum- ference is TT X 24 = 75.5 in. The slot pitch may be determined for different numbers of slots per pole per phase, as follows: Slots per pole per phase . . 1 O 3 4 Slots per pole 3 6 9 12 18 Slots 24 48 72 144 Ql/^j. .-.;.*. ^U / *"" \ : r . / ,Vioe; 31 A blot pitcn 1 , , i in incnes . 1* .57 .05 . 785 0.524 About half of the slot pitch will be required for the tooth. Considering, therefore, the insulation requirement given in Table X, it is fairly apparent that a small number of slots per pole per phase should be chosen. It will be assumed that. there are 2 slots per pole per phase. Each slot will then contain, 56 -j- = 28 conductors. A very good arrangement of conductors in a slot is that shown in Fig. 225, which permits of easy insertion of the coils. TABLE X Voltage (phase) 110 440 , 1,000 2,300 6,600 16,000 Insulation, a 20 mils 25 35 50 90 130 The size of the conductor must next be determined. As a guide for this, it may be taken as permissible to use current densities up to 2500 amp. per sq. in. in low-voltage machines, and up to 1200 amp. per sq. in. in high- voltage machines. Assuming a density of 2000 as reasonable the required area of conductor to carry 25 amp. is X It is seldom good practice to use wire heavier than No. 10 B. & S. As 0.0125 sq. in. corresponds nearly to No. 8, it will be preferable 292 ELECTRICAL ENGINEERING to divide this area among several wires in parallel. The con- ductor used consists of four No. 14 wires in parallel, having a combined cross-section of 4 X 0.00323 = 0.01292 sq. in., and giving a resultant current density of 1925 amp. per sq. in. The arrangement of wires in the slot is similar to that of Fig. 225. There are four groups of 28 wires each, the wires being placed four abreast and seven deep. Each layer of four wires is insulated from those above and below it. Width of copper in the slot is 8 X 0.064 in. = 0.512 in. Width of insulation = 0.238 in. Width of slot = 0.512 + 0.238 = 0.75 in. Depth of copper in slot = 14 X 0.064 = 0.896 Depth of insulation = 0.59 Depth of wedge =0.2 Depth of slot = 1.686 = 1 *K6 in. Width of tooth at face = slot pitch slot width = 1.57 0.75 = 0.82 in. circumference at base no. teeth TT X (24 + 3.375) Width of tooth at base = - ^ t ee -th ' " ' 75 48 - 0.75 = 1.038 in. Flux Determination. The general equation for effective e.m.f. per phase is 10 8 where 4.44 = 4 X 6ffeCtive e " m f = 4 X -^= = 4 X 1.11 average e.m.f. 4> a total flux per pole entering the armature at no-load, t = total armature turns in series per phase, = 8 X 28 = 224, / = frequency = 60, k = constant depending on the distribution of conductors on the armature periphery. If the conductors were concentrated in a single slot per pole per phase, k would be 1. With a three-phase machine, these con- ductors would never be spread out over the entire 180 electrical space degrees of the pole pitch as in single-phase or direct-cur- rent machines, but would be restricted to one-third of this amount, or to 60, on account of the space required for the other DESIGN CONSTANTS OF ALTERNATORS 293 phases. Where there are two slots per pole per phase the e.m.fs. generated in the two slots add vectorially, as illustrated in Fig. 226, where E = EI + E 2 . k is then evidently equal to ^-^ For n slots per pole per phase, 1 o 60 2n sin 7: 2n Thus, for Supplying these values in the e.m.f. equation and solving for flux, 2300 1 * - V 0.966X60X224X4.44 = 2.3 megalines. The flux leakage factor for this machine is 1.125. .'. flux in the field at no-load is, 4> f = 2.3 X 1.125 = 2.59 megalines. . Air Gap. An approximate average value for the gap length may be obtained by reference to Table IX. In the table is found, no-load A.T. per pole _ arm. reaction Armature reaction = 1.5 X A/2 X 25 X 28 = 1490. Substituting this value of armature reaction, no-load A.T. per pole = 2.5 X 1490 = 3725. These ampere-turns are mostly required for the gap. Assum- ing 80 per cent, for this, gap A.T. = 0.8 X 3725 = 2980. Assuming, now, a gap flux density at no-load of 40,000 lines per sq. in., and substituting in the equation, A.T. (gap) = 0.313 B X l a , where l g is the length of one gap, 2980 = 0.313 X 40,000 X 1 , 0.313 X 40,000 = ' 238 294 ELECTRICAL ENGINEERING With this value for a guide, definite values may be chosen. With alternators, it is usual to shape the pole pieces so that the generated e.m.f. may more nearly approach the sine form. The gaps chosen for this machine are : gap length in center of pole = 0.1875 in. gap length at edge (maximum) = 0.386 in. average gap length, l g , = 0.2535 in. Gap area, flux 2.3 X 10 6 __ K A = fluTdelisIty = 40,000 = 57 ' 5 Sq " m ' Armature Length. The main factor bearing on armature length is flux density in the teeth. This in turn depends upon gap area, pole pitch and pole arc. The pole pitch at the armature surface is The pole arc is usually about 0.6 X pole pitch. In this machine, the ratio P le arc _ 0*0 i '.L l~ U.OO. pole pitch Assuming pole-face area = air-gap area, length of pole piece parallel to shaft is A A7 K = 11.5 in. A g 57.5 0.53 X 9.43 5 The armature gross length may be slightly greater than this to assist in the free balancing in the field. The gross length is there- fore taken as 12 in. This length would justify the use of four J^-in. ventilating ducts, one for every 3 in. The length of lami- nations is therefore 12 in. 2 in. = 10 in. Assuming 10 per cent, loss of length due to insulation between laminations, the net armature length is l a = 10 X 0.9 = 9 in. The ratio, effective length _9_ total length = 12 = Teeth Flux Density. Allowing 10 per cent, extra for "fring- ing" of the flux entering the armature from the pole face, the average number 'of teeth under the pole is fi^S X L1 - 07 X L1 " 3 ' 5 DESIGN CONSTANTS OF ALTERNATORS 295 This number varies from moment to moment according as a slot or a tooth is in the center line of the pole. Teeth area at armature face is then, At = 3.5 X tooth width X effective length of armature. In order that this area shall carry a flux density of about 90,000 lines per sq. in. in the tooth, 1 it may be calculated on that basis. Thus, the flux entering the armature at no-load is 2.3 X 10 6 lines. 2.3 X 10 6 ' At= 90,000 = 256 sq ' m " From this value of area, the length obtained is, 256 3.5 X 0.821 8.9 in. Thus, the length of 9 in. previously obtained is quite satisfac- tory, giving, as it does, a slightly less teeth density at no-load, but, as will be seen, approximately 90,000 at full non-inductive load. Armature Resistance. All data has now been obtained that is necessary to calculate the resistance of the armature winding. The length of the mean turn may be taken as twice the gross length of the armature core plus nine times the diameter per pole; or the length of the mean turn = 2L + 9 D/pole = 2 X 12 + 9 X 3 = 51 in. -g- 4.25 ft. Total length = turns per phase X mean turn. = 8 X 28 X 4.25 = 954 ft. 2 9 Resistance of four No. 14 wires in parallel = -j- = 0.725 ohms per 1000 ft. /. R a per phase = X 0.725 = 0.69 ohm at 60C. For 25-cycle alternators 110,000 lines per sq. in. is suitable. 296 ELECTRICAL ENGINEERING Voltage drop per phase = IR a = 25 X 0.69 = 17.25 volts. The full-load e.m.f. per phase = E + IR a approximately. = 1330 H- 17.25 = 1347 volts Magnetic Circuit Dimensions. Sufficient data is now at hand to enable the making of a sketch which shall show approximately how the available space may be utilized. Fig. 227 represents such a section of the magnetic circuit. The next step is to construct a table for the condition of no-load and normal voltage, from which is obtained the total required number of field ampere-turns. Some of the data in this table have already been obtained, especially the required fluxes in the different parts. FIG. 227. The yoke is left out of consideration, its magnetic length being very small in revolving field machines of few poles. The armature and pole sectional areas are arbitrarily chosen to give appropriate densities. The length of the pole depends upon the space required by the field winding. The field values ob- tained for this machine are given in Table XI. Material of the armature core is standard sheet iron of 0.014 in. thickness. The field core is built up of thick steel punchings. DESIGN CONSTANTS OF ALTERNATORS 297 TABLE XI Magnetic data. No-load, normal voltage Part Flux (mgl.) Area B A.T. per in. Length A.T. Teeth 2.3 (face) (base) 2.3 1.15 2.59 26.0 32.8 57.5 28.2 27.5 89,000 70,000 40,000 41,000 94,200 15. 6\ 7.0/ 11 ' 3 2.8 74.2 1.6875 0.2535 6.0 6.25 19 3,150 17 464 3,650 21 3,203 17 484 Gap Arm Pole Total amp. -turns . Teeth Full 2. 34 (face) (base) 2.34 1.17 2.64 -load, i lormal v 90,500 71,200 40,700 41,700 96,000 oltage ^o}"-' Gao Arm 2.85 77.5 Pole Total amp. -turns . 3,725 To the total required ampere-turns to excite the field at full-load must be added those necessary to compensate for the armature reaction. The number 3725 is the resultant, F r , Fig. 228. The total ampere-turns on the field core, F/, must be equal to the vector difference of F r and F a , where F a is the armature am- pere-turns multiplied by the field leakage factor. For full non-inductive load, ap- proximately 1.125 r FIG. 228. F f = Supplying values already obtained, F f = \&725 2 + L125~>0490 2 = 4090 A.T. For any other power factor, say 80 per cent., the required field ampere-turns are approximated as illustrated by dotted lines in Fig. 228. Thus, F'f = MF r + 1.125fl. sin 2 + l.l25F a cos 2 = \3725 + 1677lxf sin + E Zm sin = ir + x - 304 ELECTRICAL ENGINEERING The solution of this is - T 8 r r T 9 Elm - (0-0i) . t = e * \ J e.x e xo sin 6dO l_ 4-C sn _L0 rE lm - r ~ ffl r . = e x exojex si l_ x Ezm r r -0 . , j/l i /- + - -J * x sm rf(9 + Ct .C J where R fr r \ ^ = ( --- ) = cot |8. X \3 Xo/ Substituting Z 2 = .R 2 + X 2 , and tan /3i = - and determining C from the condition that when = 0i, i = o the final solution is given by = This equation may be greatly simplified by introducing certain approximations, which, for practical considerations, do not injure the value of the results obtained. Thus, in practice E^ m lies between 2 per cent, and 10 per cent, of E m , .being smaller in larger machines. Neglecting E 2m in (118), and writing E m for E lm , (118) be- comes i = ^ ^ [ - r ^ (e ~ 9l) sin (B - 0) - e~ r * (0 ~ &l} sin (0! - 0)] (119) Equation (119) is convenient for fairly accurate work and should be used for ordinary wave determinations. Nevertheless, rough approximations may be made by further simplification. Thus, assume = 90; then sin (0 0) = cos 0. Also, assume Y w = 1. Then (119) becomes SHORT-CIRCUIT OF ALTERNATORS 305 These assumptions are more or less reasonable since, in practice, /3 lies between 85 and 90, and, in concentrated field windings, the reactance is much greater than the resistance. The condition for maximum current is when 61 = o, and 6 = TT. Then, E m - r -* . +e L,-i xo J. The value of is about 0.02 in all alternators, and ^r = 0.06. XQ XQ giving e~ - 06 = 1 approximately. Therefore the maximum current at short-circuit is E m -JL. Continuing the evaluation, - is from 0.6 to 0.8. x jp .' We = - - X 1.75 (approximately). x As an example, take an alternator which has 4 per cent, react- ance. The greatest possible current that can be obtained on short-circuit is then imax = Q-QJ_ X 1.75 = 44 times normal current. To illustrate the effects of short-circuit, three typical generators are taken as examples, as follows: Class A. Engine driven generators. Reactance, x = 12 per cent., = 0.12, resistance, r = 1 per cent., = 0.01, short-circuit current under normal no-load excitation, I s = 27, where / = full-load current. Class #. Turbo-generators. x = 0.02, r = 0.01, /, = 21. Class C. Turbo-generators with external reactance, x = 0.06 (0.02 internal, 0.04 external), r = 0.01, I 8 = 21. All three machines are taken on the percentage basis; with E m = 1, I m = 1. All are single-phase generators, or, the short- circuit may be regarded as that of one phase only, of a multiphase generator. Problem 96. From the above data calculate and plot the first few cycles (2 to 4) of armature current, voltage and power. The current may be determined from (119), the voltage from (117) in which Ez m is neglected, and the power from the funda- mental relation, p = ei, where instantaneous values are con- sidered. 20 306 ELECTRICAL ENGINEERING The following values are at once obtained: r is taken equal to r; X Q = j- = ^ 1 8 1 ~ = 0.5; -- = 0.02 6 XQ Class A Class B Class C o /~ *. \ r K . IT '0\ -^r= COt /3= ( ) j 0.0833 - 0.02 0.0633 0.5 - 0.02 0.48 0.1667 - 0.02 0.1467 2 =8^)3 = 0.998 0.9013 0.9894 = 86 20' 64 20' 81 40' The only other constant factor remaining to be supplied is 0i, the time-phase angle representing the instant of closing the switch. 61 may be taken at any desired value, and it should be considered what effects are produced with different values. For convenience of calculation, and also to work under extreme con- ditions the following values of 0i may be chosen. Class A Class B Class C 01 = - 3 40' -25 40' -8 20' 86 20' 64 20' 81 40' 41 20' 19 20' 36 40' For each value of 0i a set of three curves may be obtained, and a comparative study will then be possible, both in regard to the effect of closing the switch at a different point in the cycle and with regard to the influence of the constants of the different types of machine. In the present instance the curves for the engine driven generator (class A) are produced under the condition 0i = 3 40'. The equations, with numerical values supplied, are: i = 8.32[e-- 02( ' + - 064) sin (0 - 86 20') + -- 0833 C + - 064 > = 8.32[e~ a: sin a + ~ v ] = 8.32[a + 6] e = 6 -o.02(* + 0.064) gin = 6 -* sin p = ei = 8.32[e-- 04( * + ' 064) (sin 2 cos 86 20' - sin cos sin 86 20') + -- 1033 ^ +- 064 ) s in 0]. It is not necessary to evaluate the power equation since the product ei may be taken for each angular position. The tabula- tion is given for 360 from the instant of closing the switch. The three curves are shown in Fig. 229 for something over two cycles. Figs. 230-237 are for the other cases which have been taken. SHORT-CIRCUIT OF ALTERNATORS Tabulating: Case A. t = - 3 40'. 307 0-01 15 30 45 60 75 90 120 -3 40' 11 20' 26 20' 41 20' 56 20' 71 20' 86 20' 116 20' sin -0.064 0.1965 0.4436 0.6604 0.8323 0.9474 0.998 0.8962 a = 0-8620' -90 -75 -60 -45 -30 -15 0.0 30 sin a -1.0 -0.9659 -0.866 -0.707 -0.5 -0.2588 0.0 0.5 0(rad.) -0.064 0.198 0.459 0.721 0.982 1.244 1.507 2.03 + 0.064.... 0.0 0.262 0.523 0.785 1.046 1.308 1.571 2.094 X 0.0 0.00524 0.01046 0.0157 0.02092 0.02615 0.03142 0.04188 r x 1.0 0.9947 0.9895 0.984 0.979 0.974 0.969 0.959 y 0.0 0.0218 0.0436 0.0654 0.0872 0.109 0.131 0.1745 b = e - 1.0 0.978 0.957 0.936 0.916 0.898 0.878 0.841 a -1.0 -0.96 -0.857 -0.695 -0.49 -0.252 0.0 0.48 a + 6 O.Q 0.018 0.1 0.241 . 426 0.646 0.878 1.321 t 0.0 0.15 0.832 2.01 3.55 5.38 7.30 11.0 e -0.064 0.1953 0.439 0.65 0.815 0.923 0.967 0.86 P 0.0 . 0293 0.365 1.308 2.89 4.97 7.05 9.45 0-01 150 180 210 240 270 300 330 360 146 20' 176 20' 206 20' 236 20' 266 20' 296 20' 326 20' 356 20' sin 0.5544 0.064 -0.4436 -0.8323 -0.998 -0.8962 -0.5544 -0.064 a = 86 20' 60 90 120 150 180 210 240 270 sin a 0.866 1.0 0.866 0.5 0.0 -0.5 -0.866 -1.0 0(rad.) 2.55 3.08 3.6 4.125 4.65 5.17 5.7 6.22 + 0.064 2.614 3.144 3.664 4.189 4.714 5.234 5.764 6.284 X 0.05228 0.06288 0.07328 0.08378 0.09428 0.10468 0.11528 0.12568 " 0.949 0.939 0.929 0.9195 0.91 0.902 0.891 0.882 y 0.2175 0.262 0.305 0.349 0.393 0.436 0.48 0.524 b -e~" 0.804 0.768 0.737 0.706 0.674 0.65 0.62 0.592 a 0.822 0.939 0.805 0.46 0.0 -0.451 -0.772 -0.882 a + b 1.626 1.707 1.542 1.166 0.674 0.199 0.152 0.29 i 13.53 14.2 12.85 9.7 5.6 1.66 -1.266 -2.415 e 0.525 0.060 -0.412 -0.765 -0.908 -0.809 -0.494 -0.0565 P 7.1 0.0851 -5.3 -7.42 -5.08 -1.343 0.625 1.364 Class A. 0! = 41 20' =0.718 radian i = 8.32[e- - 02 ^-- 718 > sin(0 - 86 20') - -0-0833(0-0.7l8) s i n (_4 5 )] = 8.32[ ~ z sin(0 - 86 20') + 0.707 e~ v ] e = e~ x sin 0-01 30 60 90 120 150 180 210 41 20' 71 20' 101 20' 131 20' 161 20' 191 20' 221 20' 251 20' "* 1 0.989 0.979 0.969 0.959 0.949 0.939 0.929 sin 0.6604 0.9474 0.9805 0.7509 0.3201 -0.1965 -0.6604 -0.9474 a" -45 -15 15 45 75 105 135 165 sin a" -0.707 -0.2588 0.2588 0.707 0.9659 0.9659 0.707 0.2588 c~ x sin a" -0.707 -0.256 0.2535 0.685 0.9255 0.916 0.664 0.2405 707-w 0.707 0.677 0.648 0.621 0.595 0.568 0.543 0.521 i 0.0 3.5 7.5 10.87 12.67 12.35 10.05 6.34 e 0.6604 0.936 0.96 0.728 0.307 -0.1864 -0.62 -0.88 P 0.0 3.28 7.2 7.9 3.89 -2.3 -6.23 -5.58 308 ELECTRICAL ENGINEERING 10 FIG. 229. FIG. 230. SHORT-CIRCUIT OF ALTERNATORS 309 10 FIG. 231. Class A.0! = 86 20' = 1.50 radians. The equations (423) and (421) become: i = 8.32-- 02 ^ - J - 5 ) sin (6 - 86 20') = 8.32e- x sin a, e = -0-2(0 - 1.6) sin e = -x sin Q Tabulating : 9 - 0i 30 60 90 120 150 180 210 86 20' 116 20' 146 20' 176 20' 206 20' 236 20' 266 20' 296 20' sin 0.998 . 8962 0.5544 0.064 -0.4436 -0.8323 -0.998 -0.8962 a 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 sin a 0.0 0.5 0.866 1.0 0.866 0.5 0.0 -0.5 0(rad.) 1.5 2.03 2.55 3.07 3.60 4.12 4.64 5.17 - 1.5 0.0 0.53 1.05 1.57 2.10 2.62 3.14 3.67 X 0.0 0.0106 0.021 0.0314 0.042 0.0524 0.0628 0.0734 e~ x 1.0 0.989 0.979 0.969 0.959 0.949 0.939 0.929 e~ x sin a 0.0 0.4945 0.848 0.969 0.831 0.4745 0.0 -0.4645 I 0.0 4.11 7.05 8.06 6.92 3.94 0.0 -3.86 e 0.998 0.886 0.542 0.062 -0.425 -0.79 -0.937 -0.833 P 0.0 3.64 3.82 0.50 -2.94 -3.12 0.0 3.22 e - 0i 240 270 300 330 360 390 420 450 326 20' 356 20' 386 20' 416 20' 446 20' 476 20' 506 20' 536 20' sin -0.5544 -0.064 0.4436 0.8323 0.998 0.8962 0.5544 0.064 a 240.0 270.0 300.0 330.0 360.0 390.0 420.0 450.0 sin a -0.866 -1.0 -0.866 -0.5 0.0 0.5 0.866 1.0 0(rad.) 5.69 6.21 6.74 7.26 7.78 8.30 8.83 9.35 - 1.5 4.19 4.71 5.24 5.76 6.28 6.80 7.33 7.85 X . 0838 0.0942 0.1048 0.1152 0.1256 0.136 0.1466 0.157 e~ x 0.9194 0.91 0.902 0.89 0.881 0.872 0.864 0.855 t~ x sin a -0.796 -0.91 -0.782 -0.445 0.0 0.436 0.749 0.855 t -6.63 -7.57 -6.51 -3.74 0.0 3.63 6.24 7.11 e -0.51 -0.0582 0.40 ^0.741 0.880 0.782 0.479 0.0547 P 3.38 0.441 -2.61 -2.77 0.0 2.84 2.99 0.39 ELECTRICAL ENGINEERING FIG. 232. Class B.--e l = - 25 40' = - 0.445 radian i = 45.1[e-- 02 <' + - 445 > sin(0 - 64 20') + 6 -0.5(0 +0.445) j = 45.1[e- x sin (e - 64 20') + -*'] c -0.02? + 0.445) gi sin 6 e - 61 210 4 20' 0.0756 0. -60.0 -0. -0.856 0.265 0.767 -4.015 0.0748 -0.30 34 20' 0.564 0.979 -30.0 -0.5 -0.4895 0.525 0.592 4.625 0.552 2.55 124 20' 0.8258 0.949 60.0 0.866 0.821 1.31 0.268 49.1 0.784 38.5 154 20' 0.4331 0.939 FIG. 233. SHORT-CIRCUIT OF ALTERNATORS 311 Class B.Oi = 19 20' = 0.336 radian t = 45.1[e-- 02 <* - - 336 > sin ($ - 6420') - e" - 5 ^ ~ - 336 > sin(-45)] = 45.1[e- x sin (6 - 64 20') + 0.707 e~^] - 0.336) = - 0-01 30 60 90 120 150 180 210 19 20' 49 20' 79 20' 109 20' 139 20' 169 20' 199 20' 229 20' sin 0.3311 0.7585 0.9827 0.9436 0.6517 0.1851 - 0.3311 -0.7587 -x 1.0 0.989 0.979 0.969 0.959 0.949 0.939 0.929 t"" 1 sin a" -0.707 -0.256 0.2535 0.685 0.9255 0.916 0.664 0.2405 0.7076-2" 0.707 0.542 0.419 0.322 0.2475 0.1895 0.147 0.1118 t 0.0 12.9 30.35 45.4 53.0 49.9 36.6 15.9 e 0.3311 0.75 0.962 0.914 0.625 0.1758 - 0.3108 -0.705 P 0.0 9.67 29.2 41.5 33.1 8.77 -11.39 -11.21 FIG. 234. Class B.6i = 64 20' t = 45.1e-- 02 ^ - 1.12 radian 2 ) sin (6 - 64 20') = 45.1 e- x ,sin (e - 64 20') = 45.1 <-~ x sin a' e = c -o.02(0-i.i2) gn = -x sn e - 0i 30 60 90 120 150 180 210 64 20' 94 20' 124 20' 154 20' 184 20' 214 20' 244 20' 274 20' sin 9 0.9013 0.9971 0.8258 0.4331 -0.0756 -0.564 -0.9013 -0.9971 t~ x 1.0 0.989 0.979 0.969 0.959 0.949 0.939 0.929 t~ x sin o' 0.0 0.4945 0.848 0.969 0.831 0.4745 0.0 -0.4645 i 0.0 22.3 38.25 43.7 37.5 21.4 0.0 -20.95 e 0.9013 0.985 0.809 0.42 -0.0725 -0.535 -0.846 -0.925 P 0.0 22.0 30.95 18.75 -2.72 -11.45 0.0 + 19.4 312 ELECTRICAL ENGINEERING FIG. 235. Class C.Bi = - 8 20' = - 0.145 radian i = 16.5[e- - 02 ^ + - 145 > sin (0 - 81 40') + 6 ~ 0-1667(0 + o.i45)j = 16.5[ e - * sin (0 - 81 40') + e~ V] e = e- 0.02(0 + 0.145) sin e = - x sin 0> 0-01 30 60 90 120 150 180 210 -8 20' 21 40' 51 40' 81 40' 111 40' 141 40' 171 40' 201 40' sin -0.1449 0.3692 0.7844 0.9894 0.9293 0.6202 0.1449 -0.3692 -* 1.0 0.989 0.979 0.969 0.959 0.949 0.939 0.929 a" -90.0 -60.0 -30.0 0.0 30.0 60.0 90.0 120.0 e-* sin a" -1.0 -0.856 -0.49 0.0 0.48 0.821 0.939 0.804 y" 0.0 0.0883 0.175 0.262 0.35 0.437 0.523 0.611 t -v> 1.0 0.915 0.84 0.769 0.705 0.647 0.593 0.543 i 0.0 0.973 5.77 12.7 19.55 24.2 25.3 22.2 e -0.1449 0.365 0.767 0.959 0.89 0.589 0.136 -0.3427 P 0.0 0.355 4.425 12.18 17.4 14.25 3.44 -7.61 FIG. 236. SHORT-CIRCUIT OF ALTERNATORS 313 Class C.6i = 36 40' = 0.637 radian t = 16.5[e-- 02 ('-- 637 >sin (0 - 8140')- 6-- 1667 ^-- 637 ) S in (- 45)] = 16.5[e-* sin(0 - 81 40') + 0.707 c~v"] e = c -o.02(*-o.637) sin e = -x sin ^ e - Oi 30 60 90 120 150 180 210 36 40' 66 40' 96 40' 126 40' 156 40' 186 40' 216 40' 246 40' sin 6 0.5972 0.9182 0.9932 0.8021 0.3961 -0.1161 -0.5972 -0.9182 r 9 1.0 0.989 0.979 0.969 0.959 0.949 0.939 0.929 e~ x sin a" -0.707 -0.256 0.254 0.685 0.926 0.916 0.664 0.241 Q.7Q7 tne average power during this short interval is 90 X 25 = 2250 kw., which is furnished by the destruction of the flux. SHORT-CIRCUIT OF ALTERNATORS 315 The total heat developed is i 2 Rt, or W = CpRdt = Ci 2 RdO, where ti and 0i are used to designate the initial moment of short- circuit, and t and any subsequent moment. If, for instance, 6 0i is made equal to 2?m, w is the heat generated in n cycles. The complete expression for the heat developed is obtained as follows. From (119), w = fi*Rdd = R^f f) 'yV 2 "'^' sin' (0-0) _&-!(-*) sin (0 - 0) sin (0! - 0) + e - 2 <-*> gin 2 (0! - p)]dO, where is written for > a for - and i for + - #o a; XQ x Carrying out the integration, this becomes sin 2(0 - 0) - q cos 2(0 - 0) , 2 - The maximum heat is produced when the short-circuit occurs at such a time that sin (0i /3) = 1. The maximum heat produced in n cycles is then: W = PRdO = R ^ - + - ^ , \ a? Z/ L 4a 2 .. , ai 2 (1 - 2 ai )J approx. (120) The average power developed during n cycles is E I Since the rated power of an alternator is m > the ratio Power during short-circuit _ 2^n ^ ^ W = *r av , rated power E m l m E m l m irn 2 1? T 1 On the percentage basis, ^ m = ^ or if E7 = 1, where effect- ive values of voltage and current are used, the ratio becomes ^ 316 ELECTRICAL ENGINEERING Problem 97. Calculate and plot the ratio of average power, under the condition of maximum heat (Eq. 120), to rated power, for values of n from 1 to 10, for the three classes of alternators. Class A. From the previous calculation (page 315), f) 2 = - 01 ( 8 - 32 ) 2 = - 692 0.692 0.11 a = 0.02, a = 0.0833, ai = 0.1033, ! 2 = 0.0107, 4a = 0.08, 2a = 0.1667, 2! = 0.2067, - ax 2 = 1.0107, 47ra = 0.2515 4 = 1.048 27rai =0.65 2- - 0.2043. Supplying these values (121) becomes _ -1.408/ 0.11 /I - -0-2515n = n \ " ^ 0.08 0.1667 - 0.2043(1 - 2.01254 n 1*375 -0.2515n c n 0.66 _ 0.02246 _ . 65n n = a b c -}- d. Tabulating : n 1 2 3 4 5 6 8 10 2.01254 2ni occ/i IrtAAOT OATAQO OKAO1 A OA HOK1 000 SA O OOK1 K'T 0=3 n .UI^5O4 .UVoZf . o7Uoo . oUo!4 .4UZ51 . ooo4.<2 . ZOluf 0. 20125 1.375 1.375 0.6875 0.4583 0.3438 0.275 0.2297 0.1719 0.1375 n 0.66 n 0.66 0.33 0.22 0.165 0.132 0.11 0.0825 0.066 0.02246 Onoo/m f\ /\i i OO OAAT/lfi OAAKftO Of\f\A >ino Of\f\O*7A OAAOO1 OAAOO>t A n . UZZ4O U.Ull^o .UO749 . UUOD i . UU44y 2 .UUo74 .UU2ol . 002240 0.2515n 0.2515 0.503 0.7545 1.006 1.2575 1.509 2.012 2.515 e -o. 2 s, 8n 0.778 0.605 0.47 0.364 0.284 0.22 0.134 0.081 1.048n 1.048 2.096 3.144 4.192 5.24 6.288 8.384 10.48 -1.04 8n 0.35 0.124 0.043 0.015 0.0058 0.0016 0.0 0.0 0.65n 0.65 1.3 1.95 2.6 3.25 3.9 5.2 6.5 -0..5n 0.523 0.272 0.142 0.074 0.038 0.019 0.006 0.0012 b 1.07 0.416 0.227 0.125 0.078 0.0505 0.023 0.01115 C 0.231 0.041 0.00945 . 00247 0.000765 0.000176 0.0 0.0 d 0.01175 0.003055 0.001063 0.000416 0.0001707 0.000071 0.000017 0.0000027 Pa,. 0.7233 0.5523 0.4354 0.3761 0.3239 0.2848 . 2286 0.1901 Ratio 1.4466 1 . 1046 . 8708 0.7522 0.6478 . 5696 0.4572 . 3802 Class B. a = 0.02, a = 0.5, i = 0.52, a! 2 = 0.271, = 0.01 (45.1) 2 = 20.34 20.34 = 3.24 2irn n 4 = 0.08, 2* = 1.0, 2i = 1.04, i 2 = 1.271, 47ra = 0.2515. 4ira = 6.28. = 3.267. = 0.817. SHORT-CIRCUIT OF ALTERNATORS 317 Supplying values (121) becomes ' m ' = "^TL (U)8~~ ~T 41.09 40.5 _ . 2515n 3.24 * 28n . 2.65 _ 3 . 267n - 0.817(1 - n n = a ' _ 5' - c ' Tabulating: n 1 2 3 4 5 6 8 10 a' 41.09 20.54 13.7 10.27 8.22 6.85 5.135 4.109 40.5 n 40.5 20.25 13.5 10.125 8.1 6.75 5.063 4.05 3.24 3.24 1.62 1.08 0.81 0.648 0.54 0.405 0.324 n 2.65 2.65 1.325 0.883 0.663 0.53 0.442 0.331 0.265 n c 0.2516n 0.778 0.605 0.47 0.364 0.284 0.22 0.134 0.081 c -6.28n 0.0016 0.0 0.0 0.0 0.0 0.0 0.0 0.0 g 3.267n 0.038 0.0017 0.0 0.0 0.0 0.0 0.0 0.0 6' 31.53 12.25 6.34 ^.688 2.3 1.485 0.678 0.328 c' 0.00518 0.0 0.0 0.0 0.0 0.0 0.0 0.0 d' 0.101 0.00225 0.0 0.0 0.0 0.0 0.0 0.0 Pa,. 9.66 8.29 7.36 6.58 5.92 5.37 4.46 3.78 Ratio 19.32 16.58 14.72 13.16 11.84 10.74 8.92 7.56 Class C. (jjf y\ 2 -~ j) - 0.01(16.5) 2 = 2.7225 2.7225 0.434 = 0.02, = 0.1667, ai = 0.1867, i 2 = 0.035, 27m n 4a =0.08, 2a = 0.333, 2 ai = 0.3733, 1 + i 2 = 1.035, 47ra = 0.2515. 4 = 2.093. 27r ai = 1.173. 2ai =0.361. 1 + Supplying values (121) becomes 0.434 r L av * ~^ n 0.434 rl - e -o.25i6 0.08 6.565 5.42 _ . 2515n c n n = a" - V - c" + d". -1.173n >i 318 ELECTRICAL ENGINEERING Tabulating : n 1 2 3 4 5 6 8 10 6.565 6Cf C q ooq I CK 11 i qi q i 078 0090^ OfiCCK n .000 o . Zoo . 1OO l . olo 1 . \Jl O , o^uo . OODO 5.42 5.42 2.71 1.807 1.355 1.084 0.903 0.677 0.542 n 1.302 0K1 326 OOfiO 217 n iAq 1302 n l.oUJ . OOJL . ^UU W . \. 9 \J . XUO U . JLOU 0.157 0.157 0.078 0.052 0.039 0.031 0.026 0.0195 0.0157 n ^-0.25157* 0.778 0.605 0.47 0.364 0.284 0.22 0.134 0.081 e -2.093n 0.124 0.015 0.0018 0.0 0.0 0.0 0.0 0.0 e -1.173n 0.308 0.097 0.029 0.0095 0.0025 0.0005 0.0 0.0 V 4.22 1.64 0.849 0.493 0.308 0.1988 0.0906 0.0439 c" 0.1615 0.00975 0.00078 0.0 0.0 0.0 0.0 0.0 d" 0.0483 0.00756 0.00151 0.0 0.0 0.0 0.0 0.0 Po,. 2.23 1.64 1.31 1.15 1.00 0.88 0.73 0.61 Ratio 4.46 3.28 2.62 2.30 2.00 1.76 1.46 1.22 20 18 16 14 12 10 xo. As an illustration of the power developed under short- circuit, consider the genera- tor of class A. The average power developed in the first cycle under the worst condi- tion is found to be 0.7233, where E m = 1, I m = 1, and P av . = normal power output 123 45 Cycles 678 10 FIG. 238. If, now, a 25-cycle machine of 5000-kw. rating of this type, is considered, the aver- age power during the first cycle is 10,000 X 0.7233 = 7,233 kw. The instantaneous maximum of power has already been found to be 95,000 kw. Problem 98. Determine the above relation for the machines of classes B and C. SHORT-CIRCUIT OF ALTERNATORS 319 Stresses on End -connections of the Armature Coils. When end-connections run parallel for some distance, the forces exerted on them are often very great at the instants of heavy current during short-circuit. The force at any time may be determined to a sufficient degree of approximation by multiplying the average density of the flux through one conductor due to the other conductor, by the current in that conductor. Consider two similar conductors of radius, r, with a distance, d, between centers. To find the average flux through conductor B due to the current in conductor A . The flux through any ele- ment, dx, of B is, per centimeter length of conductor, 2Idx nut ~ j 2-n-X X where I is in abamperes, and /* is taken as unity. The average flux density is then: h ' efo /, d + r r In general, the force exerted is BIl dynes, where I is length of the wire in centimeters VI .76" 1 FIG. 239. FIG. 240. MM, t F /2 1 d + r j 1 nus, force per cm. = -y = log -j - - dynes. t T d T ... , F P , d+r [f 7 is in amperes, y == log If dimensions are given in inches, the formula remains the same. Example. Consider two adjacent conductors, as shown in Fig. 240. The area of each conductor is 0.2345 sq. in. The current density is taken as 2000 amp. per sq. in. under normal conditions. Therefore maximum normal current is I m = \/2 X 0.2345 X 2000 = 664 amp, d = 0.5, r = 0.1562, I = 20. 320 ' ELECTRICAL ENGINEERING The maximum force under normal load is then (664) 2 X 20 0.6562 366,000 F = 100 X 0.1562 log 03438 = 366 > 00 dynes = 44^000 = 0.822 Ib. This shows that the force under normal conditions is very slight. Under short-circuit the ratios of maximum current to normal current for the three classes of machines considered, were, respectively: for class A, 14.2 for class B, 52.0 for class C, 25.3. Thus, the maximum short-circuit forces are, for the three classes under the dimensions assumed: F A (max.) = 0.822 X 1O 2 = 166 Ib. F B (max.) = 0.822 X 52^_ = 2220 Ib. F c (max.) = 0.822 X 25.3 2 = 527 Ib. Problem 99. Discuss the effects of changing the values of r and d on the forces exerted on the end-connections. In general, the effect of short-circuit as obtained in machines of class B, was much decreased by the addition of external reactance, as exemplified in class C. What change in the relative positions of the end-connections would be necessary to reduce the force as obtained for class B to that of class C machines? Multiphase Short-circuits. The voltage of any phase, m, of a multiphase, alternator in the steady period of operation is ex- pressed by _ . / . 27TW e m = E m sin Thus, for a three-phase generator, the voltages are e\ = E m sin (ut + 0) 6 2 = E m sin (ut + 120) 6 3 = E m sin (co + 240) where m has the values, 0, 1, and 2, respectively. (For two-phase alternators, n must be taken as 4, not as 2, since the voltages differ by 90, not by 180.) The currents of a three-phase alternator are: ii = I m sin (coZ + 0) it = I m sin M + 120 + 0) it = I m sin (co* + 240 + 0). SHORT-CIRCUIT OF ALTERNATORS 321 The transient voltage, for example, of the second phase, is, from (117) in which E m is substituted for Ei m , and E 2 is neglected, as in the later calculations, 6 2 = E m sin(6 + 120)~^ ( ' 9l \ Equating this to i z r + x ~i as previously done for the single- phase machine, the current during the transient period is found to be ' sin (0 + 120 - ft - ' sin (0i + 120 -ft]- A still shorter but less close approximation is made by con- X sidering - = 1, and = 90. The current is then 12 = cos 120 ) - f-% (e - 9 * cos (B + 120)]. In a polyphase generator, the current for any phase is given approximately by E m r --(o-ei) ( n . -.- / im = [e * ; cos^i+-^- -e * cos+^- (122) where n is the number of phases and m has the values 0, 1, 2, . (n - 1). Power developed in any phase, at any instant, is the product ei. The whole power of a three-phase generator is, at any instant, the sum of the three products, e&i, e 2 iz, e 3 i 3 , of the individual phases. Problem 100. Perform the operation just indicated and prove that the power of a three-phase generator is sin- (- This equation shows that power of a polyphase generator is entirely independent of the time of closing of the switch. This time may have any value assigned to 81, but the time at any in- stant after the switch is closed is represented by B 0i, which is independent of 0i. This is quite different from the case of single-phase short-cir- cuits in which the power, similarly determined, is 21 322 ELECTRICAL ENGINEERING E P (one-phase) = cos Oi sin d-Q.5e~ 2 x<> (0 ~ ei} sin 20] In this equation, enters independently of lt Cos 0i is, of course, a constant. From the power equations, the torque on the shaft at any instant may be determined. Problem 101. Show that the maximum power of a single-phase short- circuit on a three-phase machine is two-thirds of that of a three-phase short-circuit on the same machine and explain in words the basis for this relationship. Armature Reaction. For a three-phase generator in the steady state of operation, the armature reactions of the three phases taken separately have been found to be: F A1 = i^T cos 0, F A2 = *V F A3 = i,T cos (e + y) = i 3 T cos (0 + I) , where T is the number of effective turns per phase and 0, + -=-> o 4rr + -o~ represent the angular space positions of the armature core with respect to the field core. Substituting the values of i from (122) the transient values of the armature reaction are: ErnT[ -'-(6-9!) -^(0-0i) /I + COS 20\ "| -- ' -- 2 -- ) \ --- Al = -- x cos 0! cos - m - g - l 2w\ / 27T\ F A2 = e s v cos i + cos 6 + - cos SHORT-CIRCUIT OF ALTERNATORS 323 Adding these three equations, the total three-phase armature reaction is: cos fl _ 9i _ Problem 102. Prove that the armature reaction of a polyphase generator is: ET rjl r~ T 7*0 ~~\ n Jl/m-i I (6 0j) . . - ((? 81) I /'1OQ'\ 2 x L Problem 103. Plot single-phase and three-phase armature reaction curves for the alternator for which waves of e, i, and p have been derived, and discuss their characteristic differences. Electromotive Force and Current Induced in the Field Windings. Excessive voltage may be induced in the field windings and cause breakdown of insulation. In general, the induced voltage is proportional to ^7 It is, however, difficult to obtain a reliable value of the voltage owing to the fact that the flux cannot pene- trate uniformly into the magnet cores during the exceedingly short time allowed by the rapidly changing current. The induced field current may also be abnormally great. By installing a circuit breaker in the exciter circuit, the rush of cur- rent may cause the circuit to be opened, thus taking off the field current from the short-circuited alternator. Example. Let the normal field excitation be 18,000 amp.- turns per pole, and the normal armature reaction be 9000 amp.- turns. If the armature reactance is 10 per cent., the maximum short-circuit current would be approximately seventeen times normal current. The armature short-circuit amp.-turns are then 153,000. Assuming 20 per cent, leakage between armature and field, the effective armature reaction is 0.8 X 153,000 = 122,000 amp.-turns on the field core. The field current may then attain the value of 1 22 000 ' X normal = 6.8 X normal current. If the circuit breaker is set for twice normal current, it will open the circuit. CHAPTER XLI SYNCHRONOUS MOTORS When the ordinary alternator is supplied with electrical energy and made to do mechanical work, it becomes a synchronous motor. The name is meant to indicate its chief characteristic, namely that of running in exact synchronism with the generator which supplies it with energy. If the frequency of the generator is 60 cycles per second, that of the motor its counter e.m.f. is also 60 cycles. This condition is the result of the electromag- netic relationship between the field and armature cores; the field core changes its position in space by means of mechanical rota- tion, the position of the magnetic field due to the armature magnetomotive force changes in space because of the time-phase relationships and alternation of the currents. The driving force of the motor is maintained only by the existence of a constant relationship between the field and armature m.m.f. The rate of rotation of the armature m.m.f. is fixed by the frequency of supply. The field has no fixed rate of rotation of its own and is therefore free to accept that imposed by the armature. The operation of the synchronous motor may be affected either by changing its load or by altering its field excitation. These may be called primary means of adjustment since they are ap- plicable to any motor in operation. Since, however, the speed cannot be changed, it becomes a matter of great interest and also of importance to find out what is changed, and what peculiar and valuable characteristics are associated with this hitherto un- encountered characteristic of synchronous speed. There are also secondary means of adjustment by which varia- tion in the motor performance may be brought about. These involve changes in the constants of the line or the motor circuit. Thus, in the matter of design, it is important to study the effects of different values of resistance and reactance of the armature. In operation, with a constant generator terminal e.m.f., resistance and reactance may be inserted or withdrawn from the line, thus altering the total r and x of the circuit. 324 SYNCHRONOUS MOTORS 325 A thorough understanding of the effect of these constants is essential from a practical as well as a theoretical point of view. A motor which, for instance, operates perfectly satisfactorily on one line may be entirely unstable and even unable to carry its load or even a small fraction thereof on another line. It will, for instance, be evident that a high resistance line tends to make the motor unstable unless the reac- tance is also considerable. In synchro- nous motor operation a fair amount of line reactance is essential; in fact, the very ability of the motor to carry load depends upon the presence of reactance in the motor circuit. Let E be the e.m.f. counter generated in the motor. The resultant flux will then 'be 90 ahead of E. Assuming a current I, as shown in Fig. 241, this current produces a m.m.f. in phase with itself and which may be taken equal to it, by choosing a suitable scale. The armature m.m.f. thus produced, when added vectorially to the field m.m.f., will produce the resultant m.m.f., which gives the resultant flux r . \ Tf 2 P _ HIQ /-C/0 f Imin. = 7^; -y^, " -' Assuming now values of 7, beginning with 7 m n ., cos 5 and sin 5, and finally E, may be obtained and tabulated for each value chosen. As 5 may be either plus or minus, both values must be taken. Under the conditions of test, E is not known, but the values of the field excitation or the field current which is proportional to it are known. Having just obtained E by calculation, the field excitation, F/, is next determined as was done for the case of the generator (page 225). In this case, E terminal was the zero vector, and F/ was found to be - bC - mi) 2 + (aC - SYNCHRONOUS MOTORS 337 where a = e + ir i\x b = ix +iir F C = -jjf = 1 (for convenience) rr m == - = 0.5 (for convenience). Problem 107. Determine and plot the phase characteristics for the motor problem for the three conditions : A. Eo = 1.1, E = 1 B. #o = 1, E = 1 C. Eo = 0.9, E = 1 when P = 1, r = 0.1, x = 0.3. Solution. The curve between e and / is first obtained from the equation E = V(7r - Eo cos 5) 2 + (7z - E sin 5) 2 by substituting values of /, for which sin 5 and cos 5 can be determined, and solving for E. This curve, E vs. /, is plotted. i Next, the field excitation is obtained from equation F f = V(- bC - mi) 2 + (aC - mij* or, more simply, F f = (- mi)* + (CE - since we deal directly with induced e.m.f. as the zero vector, and not with the terminal voltage and IZ drops. The curve F/ vs. 7 is then plotted. The data given are: Eo = (A) 1.1, (B) 1, (C) 0.9, P = 1, r = 0.1, x = 0.3, C = 1, m = 0.5, i = I cos 5, i\ = I sin 5. To get minimum current, cos 5 = 1, sin 5=0. ( ^ r Ep_ lEp* _P 1.1 Jl.212 1 (A) i n in. - 2r \ 4r z r ~ o.2 \ 0.04 0.1 = 5.5 - V30.3 - 10 = 5.5 - 4.5 = 1. (B) 7^. = ^2 - -4^-10 = 5 - 3.87 = 1.13. (C) /. = - i - 10 = 4.5 - 3.24 = 1.3 Also, cos 5 = T = ; sin 5 = \/l - cos 2 5 (A) cos , = jj = - + 0.09097 (B)cos5 = j +0.17 (C)cosS = i^ +0.1117. 338 ELECTRICAL ENGINEERING Tabulating, Case (A): / 1.0 1.1 1.3 1.5 1.8 2.5 4.0 Jr 0.1 0.11 0.13 0.15 0.18 0.25 0.4 Ix 0.3 0.33 0.39 0.45 0.54 0.75 1.2 0.909 0.909 826 0.7 0.605 0.505 0.363 0.227 I 0.909J 0.0909 0.1 0.1181 0.1363 0.1637 0.2272 0.3633 Cos 5 1.0 0.926 0.8181 0.7413 . 6687 0.5902 0.5903 Sin 5 0.0 0.374 0.574 0.67 0.744 0.805 0.805 Eo cos & 1.1 1.02 0.9 0.815 0.735 0.65 0.65 Eo sin S 0.0 + 411 0.631 0.737 0.818 0.885 0.885 Ir Eo cos 5 -1.0 -0.91 -0.77 -0.665 -0.555 -0.4 -0.25 (Jr - Eo cos 5)2 1.0 0.83 0.593 0.442 0.308 0.16 0.0625 Ix Eo sin S 0.3 -0.081 -0.241 -0.287 -0.278 -0.135 +0.315 Where E sin 5 is + (Ix - Eo sin 5)2 0.09 0.00657 0.0581 0.0875 0.0773 0.01825 0.0994 E* 1.09 0.83657 0.6511 0.5245 0.3853 0.17825 0.1619 E 1.042 0.914 0.807 0.724 0.62 0.422 0.402 Lagging current. Ix Eo sin d 0.3 0.741 1.021 1.187 1.358 1.635 2.005 Where Eo sin 6 is . (Ix - Eo sin 5)2 0.09 0.55 1.045 1.412 1.845 2.68 4.02 tf 1.09 1.38 . 1.638 1.854 2.153 2.84 4.0825 E 1.042 1.172 1.278 1.36 1.466 1.681 2.02 Leading current. i 1.0 1.02 1.064 1.113 1.204 1.478 2.362 ii 0.0 0.411 0.745 1.005 1.34 2.013 3.22 mi 0.5 0.51 0.532 0.556 0.602 0.739 1.181 -mi) 2 0.25 0.26 0.284 0.31 0.363 0.547 1.4 mil 0.0 0.205 0.372 0.502 0.67 1.006 1.61 (CE - mil) 1.042 0.709 0.435 0.222 -0.05 -0.584 -1.208 Lagging. (CE - mil) 2 .09 0.503 0.19 0.0494 0.0025 0.341 1.46 Lagging. (CE - mil) .042 1.377 1.650 1.862 2.136 2.687 3.63 Leading. (CE - mil) 2 .09 1.9 2.73 3.47 4.56 7.25 13.2 Leading. F/2 .34 0.763 0.474 0.3594 0.3655 0.888 2.86 Lagging. F/ .34 2.16 3.014 3.78 4.923 7.797 14.6 Leading. Ff .157 0.873 0.688 0.6 0.605 0.941 1.69 Lagging. Ft .157 1.47 1.735 1.94 2.218 2.788 3.82 Leading. Case B: I 1.13 1.2 1.4 1.8 2.5 4.0 Ir 0.113 0.12 0.14 0.18 0.25 0.4 Ix 0.339 0.36 0.42 0.54 0.75 1.2 1 I 0.885 0.833 0.714 0.555 0.4 0.25 0.1J 0.113 0.12 0.14 0.18 0.25 0.4 Cos 5 1.00 0.953 0.854 0.735 0.65 0.65 Ir Eo COB 6 -0.887 -0.833 -0.714 -0.555 -0.4 -0.25 (Ir- Eo cos 5)2 0.788 0.695 0.51 0.308 0.16 0.0625 Sin 5 0.0 0.303 0.520 0.678 0.760 0.760 Ix Eo sin 5 0.339 0.057 -0.10 -0.138 -0.01 0.44 Where sin 5 ia + . (Ix - Eo sin 5)2 0.115 0.00325 0.01 0.0191 0.0001 0.194 E* 0.903 0.69825 0.52 0.3271 0.1601 0.2565 E 0.95 0.835 0.72 0.572 0.40 0.5065 Lagging Ix Eo sin 5 0.663 0.94 1.218 1.51 1.96 Where 5 is SYNCHRONOUS MOTORS Case B: (Continued) 339 (Ix Eo sin S) 2 0.44 0.887 1.488 2.285 3.85 E* 1.135 1.397 1.796 2.445 3.9125 E 1.064 1.18 1.34 1.56 1.977 Leading. i 1.13 1.145 1.196 1.324 1.625 2.6 mi 0.565 . 5725 0.598 0.662 0.8125 1.3 (-mtV 0.32 0.328 0.359 0.44 0.66 1.69 ii 0.0 0.3635 0.728 1.22 1.9 3.04 mil 0.0 0.18175 0.364 0.61 0.95 1.52 CE - mil 0.95 0.653 0.356 -0.038 -0.55 -1.014 Lagging. (CE - mil) 2 0.903 0.426 0.127 0.001445 0.303 1.03 F/ 2 1.223 0.754 0.486 0.441 0.963 2.72 Ff 1.105 0.868 0.697 0.664 0.981 1.648 Lagging. CE mil 1.246 1.544 1.95 2.51 3.50 Treacling. (CE mil) 2 1.558 2.39 3.8 6.3 12.25 F/ 2 1.886 2.749 4.24 6.96 13.94 Ff 1.373 1.657 2.057 2.64 3.73 Leading. Case C: 7 1.3 1.35 1.5 1.8 2.5 4.0 Ir 0.13 0.135 0.15 0.18 0.25 0.4 Ix 0.39 0.405 0.45 0.54 0.75 1.2 1.11 0.854 0.823 0.74 0.617 0.444 0.2775 0.1117 . 1444 0.15 0.1665 0.2 0.2775 0.444 Cos S 1.0 0.973 0.9065 0.817 0.7215 0.7215 Eo cos 5 0.9 ' 0.876 0.816 0.735 0.65 0.65 Ir Eo cos 5 -0.77 -0.741 -0.666 -0.555 -0.4 -0.25 (Ir- Eo cos S) 2 0.593 0.55 0.445 0.308 0.16, 0.0625 Sin 5 0.0 0.2306 0.4222 0.5767 0.6925 0.6925 Eo sin 5 0.0 0.2078 0.38 0.519 0.623 0.623 Ix Eo sin 5 0.39 0.1972 0.07 0.021 0.127 0.577 Where sin 8 is +. (Ix Eo sin 5) 2 0.152 0.039 0.0049 0.00044 0.0161 0.333 E* 0.745 0.589 0.4499 0.30844 0.1761 0.3955 E 0.8626 0.767 0.67 0.555 0.419 0.629 Lagging. Ix Eo sin 5 0.6128 0.83 1.059 1.373 1.823 Where sin 8 is -. (Ix Eo sin 5) 2 0.376 0.69 1,12 1.89 3.33 #2 0.926 1.135 1.428 2.05 3.3925 S 0.961 1.065 1.193 1.43 1.84 Leading. i 1.3 1.315 1.36 1.472 1.805 2.89 mi 0.65 0.6575 0.68 0.736 0.9025 1.445 (-mi) 2 0.423 0.4325 0.463 0.542 0.815 2.09 ii 0.0 0.3115 0.633 1 . 039 1.73 2.77 mti 0.0 0.15575 0.3165 0.5195 0.865 1.385 CE mil 0.8625 0.6113 0.3535 0.0355 -0.446 -0.756 Lagging. (CE - mil) 2 0.745 0.375 0.125 0.00126 0.199 0.572 F/ 2 1.168 0.8075 0.588 0.54326 1.014 2.662 F/ 1.08 0.898 0.766 0.736 1.007 1.63 Lagging. CE - mil 1.117 1.382 1.7125 2.295 3.225 Leading. (CE - mil)'' 1.25 1.915 2.945 5.29 10.4 F/ 2 1 . 6825 2.378 3.487 6.105 12.49 Ff 1.297 1.54 1.865 2.472 3.535 Leading. 340 ELECTRICAL ENGINEERING 7 3.5 7 7 7 3.0 2.5 7 2.0 z 1.5 1.0 0.25 0.5 0.75 1.0 1.25 Induced E.M.F., E FIG. 251. 1.5 1.75 0.5 1.0 1.5 2.0 2.5 Field Excitation, F f FIG. 252. 3.0 3.5 4.0 SYNCHRONOUS MOTORS 341 Phase characteristic curves are shown in Figs. 251 and 252 for the three cases considered. It is also of interest to see how the phase characteristic is af- fected by a change in the amount of the motor load. Accord- ingly curves are drawn for the second case (E Q = 1, E = 1), under 0,2 0.4 0.6 0.3 1.0 1.2 1.4 1.6 1.8 Induced E.M.F., E FIG. 253. the three conditions P = 1, P = 0.5, P = 0. These are shown in Fig. 253. These curves are sometimes called V-curves. As the friction loss is included in the load, the condition P = can never be attained. In practice, the curve obtained with the motor running light approximates to this, however. The dotted line gives the locus of the minimum current points which is also the current at unity power factor. CHAPTER XLII INDUCTION MOTORS The production of torque, and the consequent operation, of direct-current motors is readily understood since the condition of wires carrying current placed in a field at right angles to the direction of the lines of force is quite apparent in both the shunt and the series types. If alternating current is supplied to the terminals of a direct- current series motor, the motor might reasonably be expected to run. In such a case the current is the same in both the field and the armature coils and since the flux is in time-phase with the current which produces it, the condition for the production of torque is satisfied. Moreover, since the alternation of the flux and the current is simultaneous, the direction of the torque is not changed though it pulsates in value. Such a motor would have a low power factor, due to its great inductance, and low efficiency due to its great copper and core losses, the latter being excessive with unlaminated field structure. When alternating current is supplied to a shunt motor, the condition for operation is not so well met. In this case, the cur- rents in the armature and the field coils will no longer be in time- phase with each other. The current in the field coils will lag by nearly 90 time degrees behind the voltage, while that in the armature will have only a slight time lag. This difficulty might be obviated theoretically by placing a suitable condenser in series in the field circuit. Practically, however, such a condenser would be too large and expensive to warrant its use. The question then naturally arises: Why not excite the field from the other phase of a two-phase supply? The trouble with such a solution, assuming that a two-phase supply is available, is that one-phase would be loaded with a wattless current ; moreover, the armature reaction and the torque would be pulsating. A natural suggestion might be to run two motors so as to balance the phases. The next step in the development of the alternating-current motor would be to omit the commutator, applying the well-known 342 INDUCTION MOTORS 343 principle of the production of currents by induction, as is done in the transformer. In this case the field winding acts as the primary, and the armature winding as the secondary coil. Currents induced in the armature would have directions as shown by the crosses and dots in Fig. 254. This arrangement would give no resultant torque, the torque due to the upper conductors being equal and opposite to that due to the lower conductors. Therefore another set of poles (shown dotted) should be intro- duced in space quadrature to the original poles, and the flux due to these new poles should be in time-phase with the armature current. This means a quadrature relationship in both time and space between the two sets of poles, exactly as is the case of a two-phase FIG. 255. system. The resultant flux acting on the armature forms the well-known rotary magnetic field. The Rotary Field. The production of the rotary magnetic field may be considered as due to the currents in two sets of coils as shown in Fig. 255 (a) . The current in phase A sets up an alternat- ing flux through the armature in the horizontal direction, while that of phase B sets up a similar flux in the vertical direction. These fluxes have the space relationship shown in Fig. 255 (6). The time relationship of the fluxes is shown by their equations. Thus A = 3> m sin wt fa = 3> m sin ( + 90) The resultant flux at any instant will then be composed of a horizontal component having the value of <^ at that instant, and a vertical component t 1 0.866 0.5 -0.5 -0.866 -1 Sin ut j cos ut - jl 0.5 - jO.866 0.866 - jO.5 1 0.866 + jO.5 0.5 + jO.866 ft These vectors are shown plotted through 360 in Fig. Thus, the locus of the ends of the resultant flux vectors is a circle of radius w , and the speed of rotation of the flux is /, the frequency of the alternating current. Problem 108. In a similar manner, show the relationship of the fluxes in a three-phase armature, and prove that the resultant flux is a uniformly rotating vector of magnitude 1.5$. Considering, now, that such a rotating flux will be cutting the conductors of the arma- ture; if the latter is short-circuited it is evi- dent that very large currents would be induced in it. As the speed of the armature in- creases, the rate of cutting of its conductors by the flux decreases, with a consequent decrease of induced e.m.f. and current. If the armature were to run at synchronous speed, no current would be set up in its conductors, and hence there could be no torque. Theory of Operation. Assuming a 1:1 ratio of turns of the two windings, as was done in the case of transformers. FIG. 256. FIG. 257. Case 1. Armature at Standstill. Let an electromotive force be impressed upon the primary, or field, winding so as to cause the current 7 o to flow. This current sets up a flux, , which induces electromotive forces, Ei, in both windings. Since the INDUCTION MOTORS 345 secondary, or armature, is short-circuited, the current 7i flows, so that Ei is used up in overcoming the resistance, /^i, and the reactance, I&i, of the secondary. The primary current 7 as in the transformer, must be the vector sum of 7 o, the exciting current, and I\ the load component. The impressed e.m.f. E is, likewise, the sum of the IoZ Q drop and E i} that which is supplied to overcome the induced e.m.f. in the primary. Primary and secondary phase angles are given by and 61, respectively (Fig. 257). Case 2. Armature at about Halt-speed. With the same im- pressed e.m.f., EQ, the vector diagram for half-speed becomes altered, due to the reduced Ei in the secondary and the reduced secondary reactance. Assuming constant secondary inductance, LI, the secondary reactance, x\ = 2ir/iLi, is directly proportional to the difference in speed between the rotary field and the arma- ture. This difference in speed, expressed in per cent, of synchro- nous speed, is called the "slip" of the motor, and is denoted by s. Thus, at standstill s = 1, and x = x\] at half-speed, s = 0.5 and x = 0.5zi. For any speed, 1 s, x = sxi. As the speed increases, therefore, the secondary reactance becomes less important, 0i decreases, and the value of I\ is governed to a greater extent by the secondary resistance. Since Ei decreases in the secondary, 7i also tends to decrease, this tendency being counteracted in part, however, by the re- duction in reactance. The primary current, 7o and the IQZQ drop are reduced nearly in proportion to 7i. EQ being constant, the voltage ( Ei) is somewhat increased since Ei = EQ IQZ Q . Therefore 7 00 is increased, and likewise the flux . 346 ELECTRICAL ENGINEERING The entire induction motor circuit may be represented by an "equivalent circuit," as was done with the transformer, page 178, in which, however, the slip, s, enters as a factor with reference to both the secondary reactance and the load. This diagram, Fig. 259, refers to one phase only. E Q is phase voltage, and Ii 2 R is the nth part of the motor load, where n is the number of phases. The magnitudes of the various quantities are readily apparent from an inspection of the "equivalent circuit" diagram, whatever may be the load placed upon the motor. FIG. 259. Referring to the vector diagrams, Figs. 257 and 258, or to the "equivalent circuit" diagram, Fig. 259. Let I m magnetizing current, I h = core-loss current, ZQ = TQ + j XQ = primary impedance, Zi = r\ + j sxi = secondary impedance. Then and 0oo = = conductance of exciting circuit, 6 o = = susceptance of exciting circuit, where e t = primary induced e.m.f. and I m is a positive quantity so that b QQ is always negative, these quantities being all taken per phase. Also, s = slip. At standstill, s = 1; at synchronous speed, s = 0; at normal full-load, s is usually about 0.02 in per cent, of synchronous speed. Then, S6i = secondary induced e.m.f. INDUCTION MOTORS 347 Let 6i be chosen zero vector. Secondary current may be written where sr 1 1 9 I T.i T" and 2 i 9 2 The exciting current is The primary current is IQ = IQQ -\-I\- i(a\ -\- o + j(o>z + 6 o)). The e.m.f. consumed by the primary impedance is e i(bi + jbz)(r Q -\-jxo) and the impressed voltage is E Q = i + /o2 = i + e t -(6i + j6 2 ) (r + jaJo) 6 2 r )] The torque, in any motor, is proportional to the current and the flux in time-phase therewith. If 1 1 is the secondary current due to a certain phase of the pri- mary, whose induced e.m.f. is e^ then the power component of /i has been shown to be e-too oo t^o>oi> os OOCOOS-COOOrH C^OOSOSCOrHrHOSOOS g I I OOOOi-iOOOCOOCO I .8 * _ ^ .._ - . l*-OOrHOOOCOOCO ,TH^CO I I X; I I O^t^-iO lO *O 1O OSrHiO CO 00 CO COO^OOrHOOOOSQOOOOSOOrHQCOrH O O5 T-H 00 CD I ;iO O CO t^" CO (N OiO0000 C UT3 rHOSC cOI>iOOOOOCOcO. t>. 10 i-HrH^HTjH OCO i-H (N O O O ^TH 8^: X CO OS CO OS CO i i I I 00 OO l> t^ rH O OOrHOOrHOOS OS N I OOOSOrHOT^OrHrH OO' CO rHOOOOOOOrHOrHOrH I I T-I O H O O O CO ^^ 00 CO OO 00 OS I s * ^^ CO OS C^ CO ^O CO CO rH rH CO CO CO ^O 00 CO ^^ CO (N 00 SOOC5OCOrHCOCOOt>- rHrHCO^OCOO^O OO VOlOOOOOTjHCO^rHfNl^-OOC^rHHH OO^OOiOCOC<|rHCO COOOOOOOrHOrH OOSOSiOT^TtirtHOOiO-*^OOSOO^. oooooo I I O O O *O O CO CO ' OOOOOOOCOOpp COpOO' ddo'dddddddd CO I I I t> rH OO CO OO5O5 H U to co k bi b *-r Perfon L =0.02, OJ t-0.3. ffl nanc -Xr ^^. e Cu -0.12 1 -^ rves ~. X ;^-= * _ fficie i ncy > 1 X 3 ?par y tf ^ ^ ifiS^ ^^ ^^s ' \ / '\ | / ., ^ / / ^ / x / / \ V 0-& >*> ^ s^' / '/ ^s ^ 44. C^ r^ /// > < ^ _^ ^ . ^ - . ****** "^ ^ ~^. . ^~ E PWBB5 ===== == == ii - .WM^ - ^ HH ^ HH 1 .-p 1.2 1.4 1.6 percentage values. Similar sets of curves are shown for the other cases in Figs. 261 to 266 inclusive. These curves illustrate the capabilities of the various motors under normal running conditions. Of equally great interest are the curves between speed, current 23 354 ELECTRICAL ENGINEERING Induction Motor Performance Curves ~l, ro-n-0.02, o- 1-0.12, J^-0.02, JOT" 0.2. 0.6 0.8 1.0 Output, (P m _/ ) FIG. 261. 1.4 1.6 3- u | 8 1.1 C Induction Motor Performance Curves ^o"i. ?V=o.02 f 1^=0.02, J m =0.3. 1.4 FIG. 262. INDUCTION MOTORS 355 Induction Motor Performance Curves #0=1, Tb-n-O.OS, o-S 1-0.12. 7^=0.02. /m=0.3. 0.2 0.4 0.6 0.8 1.0 Output, (P m -fo) FIG. 263. 1.2 1.4 E Induction Motor Performance Curves /& =0.02, 1^=0.3. Power 0.4 0.6 0.8 Output, (P OT -/ ) FIG. 264. 356 ELECTRICAL ENGINEERING 1.1 1.0 F Induction Motor Performance Curves EQ-=I, r =o.o2, r t =o.c It, 0.02. /m=0.3. 0.4 0.6 0.8 Output, (P m .-.f ) FIG. 265. 1.0 1.2 G Induction Motor Performance Curves -o.5. r =ri=o.o -0.01. /w=0.16. 0.05 0.1 0.15 0.2 0.25 0.3 Output, (P m -/ ) FIG. 266. 0.35 0.4 0.45 INDUCTION MOTORS 357 0.2 0.2 0.4 0.6 0.8 Speed, and Current -f- 4 FIG. 267. 1.2 1.1 1.0 0.9 0.8 |0.7 0.6 0.5 0.4 0.3 0.2 0.1 7 \ \ L 0.2 0.4 0.6 0.8 1.0 Speed, and Current -f 2 FIG. 268. 1.2 1.4 358 ELECTRICAL ENGINEERING and torque, which are characteristic of starting conditions. These are shown in Figs. 267, 268, and 269 for all cases except B and D. In these two cases the curves are very nearly the same as for case A. The differences may be seen by a glance at the tabulations. Fig. 267 is of special practical interest. Here torque-speed curves are given for a number of motors which differ only in the amount of their secondary resistance. There is practically no difference in the current curves, one curve giving the current for all motors. 0.5 0.4 0.3 0.2 0.1 0.2 0.4 0.6 0.8 Speed, and Current 7- 2 FIG. 269. 1.0 A motor may be imagined as supplied with a variable secondary resistance. Suppose that it starts with r\ = 1 ohm. The torque will be 0.85, and the current will be 0.25 X 4 = 1.00 as is seen from the figure. Thus, the motor starts with full-load torque and current. When half-speed has been attained, the secondary resistance is changed by some device to 7*1 = 0.5. The motor at once is changed from operation at the point, a, to the point, 6, and the current rises from 0.6, to which it had fallen, back to the original value of 1.00. The motor now follows the second torque curve to c, then, by a change of resistance to 7*1 = 0.2, it accelerates along the curve de, the current following curve d'e'. Another change to TI 0.05 causes the motor to run along fg, the current following /' or ?= X -^n~ - 61.8 volts to starting current -\/3 371 neutral, or 107 volts between terminals. 5. Maximum output is (Fig. 260) 1.67 watts. This corre- sponds to a maximum torque of 1.82 synchronous watts at 8 per cent, slip (Fig. 267) these values being, of course, per phase. To change to horsepower, this gives 1 f\7 max. hp. = -~r = 0.00224 per phase. INDUCTION MOTORS 361 For the motor considered, max. output max. hp. = - ; - X 75 hp. normal output = g|g X 75 = 150 hp. 6. Starting torque, if normal current only is allowed, is, since torque is proportional to (voltage) 2 , /ft - o\ 2 T = f(y ' N2 X starting torque at normal voltage. = 0.0592 X 6970 = 412.5 synchronous watts, or = 0.0592 X 123.3 = 7.3 ft.-lb. 7. Maximum output under impressed voltage of 61.8 volts per phase is equal to maximum output under normal voltage, mul- tiplied by the factor r^f) = 0.0592, or maximum output at 61.8 volts = 150 hp. X 0.0592 = 8.88 hp. (b) Motor Assumed A-connected. The student may show that in this case the required values are: 1. Full-load current = 90.3 amp. Phase current = 52.2 amp. 2. Starting current = 214 amp. per phase. 3. Starting torque = 123.3 ft.-lb. 4. Voltage to give normal current at starting = 107. 5. Maximum output = 150 hp. 6. Starting torque, with normal current, = 7.3 ft.-lb. 7. Maximum output on 107 volts = 8.88 hp. Questions. Do these answers indicate that a motor built to be operated Y-connected, may be reconnected A and will then give substantially the same performance? Show that if a certain motor A-connected is intended for operation at 100 volts, if it be reconnected Y, and operated at the same voltage, the output will be reduced to J^. Discuss, also, the effect of this change on the power factor, maximum output, torque, etc. Show that if the impressed voltage is reduced by 10 per cent., the maximum output and the starting torque are reduced by about 20 per cent., the starting current by about 10 per cent., while the efficiency remains about the same and power factor is slightly improved at light loads. This question is very practical 362 ELECTRICAL ENGINEERING since a motor designed for and rated at 125 volts may often be available for operation on a 110- volt circuit. Show that such operation might not always be practical on account of the reduction in the overload range. Would it be practical to operate at 10 per cent, above normal voltage? Show that in this case the power factor will be much poorer, particularly at light loads. Show that if the primary and secondary reactance are eaeh 10 per cent., the starting current will be slightly less than five times the full-load current. From the curves that have been given, discuss the value of starting current the ratio, : r , , ? When considering the per- ' running light current formance of the motor, especially the margin in output, show that the smallest ratio for a good motor should be about 12. The running light current is substantially equal to the mag- netizing current. Many other considerations are involved in the choice or operation of induction motors. Some of these may be briefly discussed. It has been shown that a variation of the secondary (usually the rotor) resistance is accompanied by marked changes in the performance characteristics of the motor; that higher resistance means, roughly, increased starting torque but decreased normal running efficiency. For normal operation, therefore, the smallest possible secondary resistance is desirable. This is best obtained by a type of rotor winding construction known as the "squirrel cage." In this, the conductors are heavy copper bars, lightly insulated, with only one bar to a slot. The ends of the bars are connected to copper rings which thus give a completely short-circuited winding. The resistance of such a squirrel cage affair is extremely low, and the starting torque of the motor is correspondingly low. There is no op- portunity of inserting additional resistance in such a structure. For this reason, many rotors are supplied with definite windings the terminals of which may either be brought out to slip rings on the shaft, or be connected to a revolving resistance mechanism carried within the rotor spider. The type of motor to be chosen depends on the use to which it is to be put. Induction motors cannot be used as indiscrimi- nately as can direct-current motors, for example. Consider a INDUCTION MOTORS 363 motor to be used in pumping against a high hydraulic head. The squirrel cage motor would not start. There might also be diffi- culty with the wound rotor type. In this case it would be necessary to have many steps in the secondary rheostat to insure against the torque falling at any instant below the required amount. Another question of importance relates to the frequency. Assume, for example, that a 5-hp., 60-cycle, 220-volt motor is required. If it is found that there is a 5-hp., 40-cycle, 220-volt motor available, will it be practicable to utilize this machine, thus saving, perhaps, the cost of a new motor? When a 40-cycle motor is operated on the 60-cycle circuit it is evident that the magnetizing current, and consequently the flux, 40 will be reduced in amount approximately in the ratio ^~, while OU the reactance will be correspondingly increased. The motor would therefore be weak in operation, that is, it may have insufficient margin in overload range. Changing from 60 cycles to 40 cycles would have just the reverse effect. The motor would have ample capacity. Mag- netic densities might be excessive, and the power factor might be considerably poorer owing to the great increase in magnetizing current. To operate the 40-cycle motor on 60 cycles would be most 60 satisfactory jf the voltage could be increased in the ratio -^ This, however, is ordinarily impossible. It is sometimes pos- sible, however, to accomplish approximately the same result by reconnecting the windings. Suppose, for instance, that the motor is A-connected. Consider changing to Y-connection, at the same time dividing each phase into two circuits and connect- ing them in parallel. Connecting in parallel changes the required voltage from 220 to 110. Changing to Y makes the required voltage 1.73 X 110 = 190. The change of frequency alone 60 would suggest a voltage of 110 X TQ = 165. Perhaps, there- fore, the change to the condition of 190 volts, that is to parallel Y-connection, will give satisfactory results in operation. An approximate value for the power factor may be obtained as indicated in Fig. 270. A right triangle is formed, the per- pendicular sides of which are per cent, load and per cent. I m 364 ELECTRICAL ENGINEERING + per cent, reactance. The reactance is the sum of both the primary and the secondary reactance in per cent. The angle is the phase angle. The basis for this approximation is found from a study of the vector diagrams, Fig. 257. Motor and Transmission Line. When an induction motor is ^ at the end of a transmission line on which ^^^^ constant voltage is impressed, the constants * %x _iL_^ of the line should be added to those of the %Load motor windings in determining the per- FIG. 270. formance characteristics. Thus, r is the sum of the primary winding and the line resistances, and XQ is the sum of the primary winding and the line reactances. Let it be assumed that a 220-volt motor is at the end of a transmission line such that maximum output occurs when the line drop has reduced the voltage on the motor to 190. /190\ 2 The maximum output is then (oon) =0.75 times its value under normal voltage. If the performance curves at normal voltage are given, these may be changed to give approximately the performance under the conditions named by merely altering the scale of abscissae so that the maximum output shall occur at three-quarters of its former value. Motor with Auto-transformer. If a motor is used where it does not have to start under heavy load, an auto-transformer may be introduced to reduce the starting current. For instance, if the auto-transformer supplies half voltage the current will be reduced to one-half, and the volt-amperes will consequently be only one-quarter of normal starting amount. On the primary side of the auto-transformer, then, the current input will be only one-quarter of the normal starting current. Such an arrangement is advantageous from the standpoint of current, but is bad where a large starting torque is required, since the torque is reduced to one-quarter its normal starting value. Many other considerations may naturally arise in reference to the induction motor, some of which may well be studied from the standpoint of its design which is taken up in the next chapter. CHAPTER XLIII STUDY OF THE DESIGN CONSTANTS OF AN INDUCTION MOTOR 1 The primary limiting features in induction motor design are (1) the possible amount of copper per inch of periphery, and (2) the smallness of the air gap. A small air gap means a cheap motor and high efficiency. Gap lengths ordinarily run from 0.02 in. to 0.06 in., the smaller values being for small machines. A relation between these two limiting features may be found in practice; thus, for each 0.01 in. of air gap, from 100 to 150 amp.- conductors per inch of periphery may be assumed as a starting basis. Let it be required to design the following motor: 7-6-10-1200-1 lOv. Since practically all induction motors are three-phase, that feature is not indicated in the rating. If the motor were two- phase, its designation would be IQ. Otherwise, the rating indicates 6 poles, 10 hp., 1200 r.p.m. at synchronous speed, and 110 volts. Air Gap. The air gap may first be assumed in accordance with practical experience. As a safe average value, let this be chosen as 0.02 in. = l g . The number of ampere-conductors per inch of periphery, from the relation given above, will then be 2 X (100 to 150) = 200 to 300. Rotor Diameter. This, also, may be taken from experience. As a trial value, let the diameter per pole be assumed as 2.5 in. The rotor diameter is then 2.5 X 6 = 15 in. = d*. Trdi 15.04 X 3.14 Pole pitch = -- = ^ = 7.88 in., o o where di is the inside diam. of the stator. 1 It is realized of course that since this is not a book on design of electric machines, the equations used are not claimed to be accurate, but they are sufficiently accurate for the purpose in view which is to give a fairly com- plete idea of how the various constants are derived. Indeed the methods suggested have been used in the design of a large number of machines that have been built. 365 366 ELECTRICAL ENGINEERING Stator Slots per Pole. This depends primarily on the slot pitch, but must be a multiple of three. For low-voltage motors the slot pitch may be quite small, say 0.65 in. As the voltage increases the space requirements of insulation will cause an increase in the pitch. Assuming 0.65 in., the number of stator slots per pole is 0.65 and the slot pitch revised is 7.88 _ 12 Slots per pole and phase are 0.657 in. - : 3 Ampere-conductors per slot will be (200 to 300) X 0.657 = 130 to 200. The number of conductors per slot will depend for their size on whether the motor is to be A- or Y-connected. The current they must carry may be calculated on the basis of 10 kv.a. input, or 3300 volt-amp, per phase. If A-wound,' T 330 i phase = = \/3 X 30 = 52 amp. amp. conds. slot = - - = amp. If Y-wound, amp. conds. 130 to 200 Conductors per slot = - - = - ^ - = 4.3 to 6.7 amp. 60 Iphase = Iline = 52 amp. Conductors per slot = - =~ - = 2.5 to 3.8. oz Slot and Tooth Dimensions. In general, it is good practice to use if possible four coils per slot. This arrangement lends itself readily to reconnection either in series-parallel, for double vol- tage, or in series for quadruple voltage. Therefore, selecting four as the number of groups of wires per slot we also get four effect- ive conductors per slot. The next step is to design a suitable slot. The deeper the slot, the more copper per inch of periphery is possible. DESIGN CONSTANTS OF INDUCTION MOTOR 367 In a given coil, however, it is not practical to wind more than four wires on edge. Therefore, a slot similar to that shown in Fig. 271 should be chosen. This gives four wires per coil, four effective conductors per slot, or a total of sixteen wires per slot. Either A- or Y-connection may be chosen. In this case it will be the latter. The actual slot dimensions for this motor are depth = 1.15 in., width = 0.34 in. Each effective conductor consists of four No. 10 B. & S. double cotton- covered wires. Each wire has a diameter of 0.112 in. over insulation, and the copper area is 0.00815 sq. in. 52 Current density is 43^0815 = 1600 amp. per sq. in., which is a rea- sonable value. The slot insulation is about 0.04 in. in thickness, being sufficient for 440 volts. Uo -4^.22 FIG. 271. The slot opening is made as small as possible to permit con- venient insertion of the coils. Its width in this case is 0.22 in., or about two-thirds of the slot width. The tooth dimensions are: width at face = 0.43 in., width at narrowest point = 0.324 in., shown respectively at a and b in the figure. Main Flux. All data are now available to substitute into the fundamental equation E = 10 8 from which the required flux, $, may be ascertained: E = j=. = 63.5 volts to neutral, V3 / = 60 cycles, t = 2 turns per slot X 24 slots per phase = 48 turns, k = constant for winding distribution = 0.96. . 63.5 X 10 8 , lfinon ' ' * := 4.44 X 60 X 48 X 0.96 = L6 ' ( Stator Length. The cross-sectional area of the iron necessary to accommodate this flux is limited by the maximum permissible flux density in the stator teeth. For 60-cycle motors the maxi- 368 ELECTRICAL ENGINEERING mum density should not exceed 90,000 lines per sq. in.; for 25 cycles the density may reach 105,000. Higher densities are apt to cause excessive heating of the teeth. It has been shown that although the windings on the stator are stationary, the effect due to the multiphase currents in them is similar to that of a revolving field structure. With alternators, the field flux is more or less evenly distributed along the surface of the pole, that is, the density is fairly uniform, and thus the flux, when plotted, approaches rectangular shape. With the distributed windings of multiphase induction motors, the space distribution of the flux is practically that of a sine wave, as will be seen by plotting the magnetomotive forces at different points along the stator periphery. With a sinusoidal space distribution, then, the maximum flux density is ~ times the average density over the surface. The net stator length may thus be determined by assuming maximum teeth density, from the relation, N , , __ flux per pole _ ~~ max. flux density X min. teeth width per pole This equation may be used, however, more advantageously as a check in settling the final dimensions both of the teeth and the stator length. The latter may best be determined at the start, by assuming a value for magnetizing current and working through the gap. Let the magnetizing current required to send flux through the gap be assumed as 20 per cent, of full-load current, = 0.2 X 52 = 10.4 amp. Gap amp.-turns = \/2 X 1.5 X 10.4 X 8 = 176.5, since there are eight turns per pole and phase. Average gap flux density is D 176 - 5 _ O &avg. gap - Q ^3 x Q Q2 - .'. Gap area per pole, 516,000 = 18 ' 3 S * teeth width per pole = (0.43 + 0.02) X 12 = 5.4 in.; where 0.02 in. has been added to the tooth width on account of unavoidable irregularities in the stampings or laminations. 18 3 .*. gross stator length = -=-~- = 3.4 in. DESIGN CONSTANTS OF INDUCTION MOTOR 369 Experience in design would now lead one to judge this length of 3.4 in. to be too short, as compared with a rotor diameter of 15 in. Therefore it will be advisable to repeat the calculations on the basis of, say, 12 in. rotor diameter, instead of 15 in. The new slot pitch then becomes, 0.657 X ~ = 0.525 in. 10 The slot and the slot opening remain as before, but the tooth face is reduced to 0.525 - 0.22 = 0.305 in. Teeth width per pole = (0.305 + 0.02)12 = 3.9 in. I Q O .'. gross stator length = -~r = 4.7 in. o.y and effective stator length = 0.9 X 4.7 = 4.23 in. This value may now be applied to the stator teeth in order to ascertain whether the maximum flux density is satisfactory or not. Minimum width of tooth = 0.19 in. Minimum net area of teeth per pole = 0.19 X 12 X 4.23 9.65 in. p -i n (\f\r\ Therefore, maximum average flux density in teeth = n ' y.oo = 53,500 lines per sq. in., and maximum flux density in teeth = I X 53,500 = 84,000. Rotor Slots. Consider, first, the squirrel cage rotor. In choosing the number of slots it is well to insurq that there shall be no possible symmetrical arrangement of the stator and rotor slots with respect to each other. In practice it is common to take one-half the stator slots 1. 72 In this case the number will be -^ -f 1 = 37 slots. Such a choice is made to prevent the existence of so-called "dead points," or positions of the rotor from which it may not start. Slot and Tooth Dimensions. At synchronous or operating speed the frequency of reversal of the flux in the secondary core is so low, being that of the slip, that the core loss in the secondary is negligible with proper teeth dimensions. This fact permits higher densities in the teeth than are permissible in 'the primary teeth. The maximum density occurs, of course, at the base of 24 370 ELECTRICAL ENGINEERING the teeth. When the rotor diameter is not great, as in small motors, a deep slot means considerable reduction of tooth area at the base compared with the area at the top of the tooth. Usually rotor slots are fairly shallow. The dimensions in the present case are given in Fig. 272. The rotor bar is nearly square in section, and is inserted in the slot from the end. It is covered with a thin layer of paper insulation, although, since the bars are all short-circuited, even this is not essential. The opening between the teeth is narrow in order to give a large flux area in the gap. At the same time, it is maintained for the purpose of reducing the leakage flux and thereby cutting down the self-inductive reactance. Considering the dimensions, as given, the minimum net area of rotor teeth, per pole, is 07 ~ X 0.32 X 4.23 = 8.35 sq. in. 4 .'. Average maximum flux density in rotor teeth is 516,000 8.35 = 61,800 lines per sq. in. Maximum density = ^ X 61,800 = 97,000. If, instead of the squirrel cage, a wound rotor is desired, the number of slots chosen will in this case Mi be 54, which will give nine slots per pole, and three slots per pole and phase. The rotor need not necessarily have a three-phase winding; any multiphase winding with the proper number of poles will do. In fact, the squirrel cage may be regarded as a winding of many phases. Slot and tooth dimensions are given in Fig. 273. From these, minimum teeth area per pole is 0.21 X 9 X 4.23 = 8 sq. in. .'. Average flux density in the rotor teeth at base is FIG. 273. 64,500 lines per sq. in. Maximum density = 9 X 64,500 = 101,300. DESIGN CONSTANTS OF INDUCTION MOTOR 371 The remaining rotor calculations are made in essentially the same way for both the squirrel cage and the wound rotor types. The former type will therefore, alone, be worked out. The student may, for practice, apply the process to the determination of constants and performance characteristics of the motor with wound rotor. Rotor Secondary Resistance. This is difficult to calculate by the usual method applied to definite windings, but may be deduced from a well-known fact that the loss, per cubic inch of copper, is 0.79 watt at 60C. when the current density is 1000 amp. per sq. in., and this loss varies as the square of the current density. The problem, then, is to find the volume of the rotor con- ductors and the current in them. The latter may be considered first. Consider the revolving flux as represented at some given instant by a pair of poles, shown in Fig. 274. At that instant the current flows downward through the rotor bars under one pole, and upward through the bars, under the other pole. These currents are indicated by "the crosses and dots in the figure. Maxi- mum current will be in the bars under the middle of each pole where the flux is maximum. Between the poles there will -pio. 274. be little current in the bars. In general, the distribution of current in the bars around the periphery will be proportional to the sine of the space angle between the bar and the neutral axis. Since, now, there is an inductive relationship between the primary and the secondary windings, as in the transformer, if the exciting current is neglected and perfect mutual inductance is assumed, the ampere-conductors around the stator periphery must equal the ampere-conductors around the rotor periphery. But the former, denoted by I a C 3 = 288 X 52 = 15,000, since there are 288 conductors on the stator, each carrying 52 eff. amp. The effective rotor current per bar will then be 15,000 IT = = 405 amp. 372 ELECTRICAL ENGINEERING Maximum current in any bar is then, 7 r (max.) = 405 X A/2 = 572 amp. 2 Average current per bar = 7 r (av.) = 572 X - = 364 amp. In the end rings which short-circuit the bars the current will vary in amount along the circumference. The section of the ring at A, Fig. 275, will carry the most current, while at B, the current will be zero. One-quarter of the bars will send their currents through section A, or, in general, section A will contain total bars the current from 37 2 X poles In this case, then, there will be j7> = 3.08 bars supplying this current. The current through A will be 3.08 times the average current per bar, or, 3.08 X 364 = 1122 amp. The average current around the whole circumference of the ring is 1122 X - = 715 amp. and the effective current in the ring is I R = 1122 -T- A/2 = 794 amp. Volume of each rotor bar is 0.5 X 0.55 X 6.7 = 1.84 cu. in. Total volume of bars is 1.84 X 37 = 68.1 cu. in. FIG Since effective current per bar is 405 amp., and, assuming 0.79 watt per cu. in. loss at 1000 amp. per sq. in., the loss in the bars is X 6!U = 11 - 79 X (o.5 X 4 55 5 Loss in bars per phase = -5- = 39 watts. o Volume of end rings is 2 X 0.5 X TT X 10.1 = 31.75 cu. in. DESIGN CONSTANTS OF INDUCTION MOTOR 373 Effective current per ring is 794 amp. .'. Loss in the rings is 794 \ 2 X 31.75 - 63.5 watts. . Loss in rings per phase = ^- = 21.2 watts. Copper loss in rotor secondary per phase = 39 + 21.2 = 60.2 watts. This loss is PR, per phase, where / and R are certain " equivalent" values of current and resistance of the secondary. Referred to the primary on a 1 : 1 ratio basis as is usual in trans- former calculations, this loss becomes 60.2 = 7 p 2 r! = 52 2 ri. 60.2 .'. TI = 27 ' = 0.0223 ohm, which is the desired rotor (secondary) resistance per phase reduced to the primary. Although this calculation is based on two assumptions, namely; negligible exciting current, and perfect mutual in- duction between primary and secondary, both of which are false, yet the error introduced is so small as not to appreciably effect the correctness of the results. The primary resistance may now be calculated in the usual way. Length of mean turn = twice the gross length of stator + eight times diameter per pole, = 2 X 4.7 + 8 X 2.22 = 27.16 in. Total length of effective conductor per phase is 27. 16 x *5g5? = 108.64 ft. . . 1 1 (\ Resistance per 1000 ft. of 4 No. 10 wires in parallel is -~- 0.29 ohm at 60C. .'. Resistance per phase, of stator, is r = X - 29 = - 0315 ohm - Primary Leakage Reactance. This is determined in the same general manner that it was done in the case of the alternator. The facts that the air gap is uniform around the periphery, that it is comparatively very short and that the rotor teeth, instead of being opposite a sojid pole, are opposite the stator teeth, 374 ELECTRICAL ENGINEERING introduce important variations into the calculations. The leakage flux may be regarded as that which passes through and around the four slots of any pole and phase at right angles to the direction of the main flux, due to the current in those slots, to- gether with that surrounding the end-connections due to the current in them. This flux may, for convenience, be divided into a number of parts, as follows: 01 = flux which crosses the space occupied by the conductors. This space has an area, a, (Fig. 276), per centimeter length of conductor. 02 = flux crossing the area, d, between the conductors and the gap. 03 = flux in the gap but not cutting the rotor conductors. This is the so- called "zig-zag" flux, because it passes back and forth across the faces of rotor and stator teeth. 04 = flux around end-connections. 05 = belt leakage flux, due to the fact that the primary and secondary coils are not always in positions to react fully on each other. This flux has to be considered only with reference to rotors with definite phase windings. In such cases the inductance, L 5 , due to it may be taken as approximately 5 per cent, of the total primary self-induction. Calculation of inductances due to 0i and 2 give FIG. 276. LI = l 47rsn : 36 d Z/2 = 4?rsn 2 - e cm., per cm. gross length of stator, where s = slots per pole and phase, and n effective conductors per slot. Similarly, i/* where -^ is the effective magnetic conductance of the path of the flux 03, per centimeter gross length of stator. This inductance is really the sum of two inductances, L 3 = L'z + L"s, of which Z/ 3 is made up of the interlinkages per DESIGN CONSTANTS OF INDUCTION MOTOR 375 ampere of all the conductors of the phase with the flux, ' 3 , which follows the path from A to B, and L" 3 is the interlinkages of the conductors in the two inside slots with the flux < 3 ", which follows the path from C to D (Fig. 277). Magnetic conductance across a rotor slot above the conductor is h T, where the letters indicate dimensions, as given in Fig. 277, in inches. This magnetic conductance is in parallel LI FIG. 277. with a conductance of variable magnitude of the path across the gap into a stator tooth and back again. When a stator tooth is directly opposite a rotor slot, this magnetic conductance is -r- * where t is the width of a stator tooth face. When the two slots are directly opposite, the magnetic conductance of this path may be neglected. In the first case (tooth opposite slot), the magnetic conductance of the combined path is 40 In the second case, it is m k The average of these two values is m k , J_ ~n h + 80' The flux 0' 3 passes through this path twice, as the figure shows, and it also passes twice across the gap as it first enters and finally leaves the rotor. The reluctance of these latter portions of the path is --- 376 ELECTRICAL ENGINEERING .'. the total average reluctance is, approximately, 2 = 2 _ , 20_ /flfo\' ~k , * " " * : : \fj and the required magnetic conductance is - ( f -Y- p'a M/o/ In the present case, supplying values, m = 0.03, n = 0.06, fc = 0.15, h = 0.26, < = 0.305, g = 0.02 ' 2 0.04 0.5 + 0.577 + 1.91 T 0.305 and 1 0.8 = Maximum magnetic conductance of the* path from C to D occurs when the stator tooth is opposite a rotor slot. The middle of the phase is here opposite a rotor tooth. In this case the magnetic conductance is t m k n + h When the slots are opposite, the magnetic conductance is 1 0.944. , t ' m k n + h When the middle tooth of the phase is opposite a rotor slot, the magnetic conductance is ~ ; = 2.98. * I f ^^ TI A n h 40 The approximate average magnetic conductance is the aver- age of 4.65, 0.944, 2.98 and 0.944. This is DESIGN CONSTANTS OF INDUCTION MOTOR 377 Inductance due to end-connections is, as has been stated in the study of the alternator, a difficult quantity to calculate with close approximation, due to the indefiniteness of the mag- netic path in the air. In the case of induction motors with wound rotors, a fair value to assume for the flux per ampere-conductor per centimeter length of the end-connections is poles diameter For squirrel-cage induction motors about 50 per cent, should be added to this, as the end-connections of the rotor may be regarded as having negligible self-induction, which affords, consequently, an increased path for the stator flux. Since length of end-connections per coil = 8 X the flux per coil per ampere-turn is 04 = 1.5 X 8 X J- = 12 J- \ p \ p 120 A per absolute amp.-turn. p With the usual double-layer winding of the stator (primary) the magnetomotive force per coil is -77-* Therefore, the total flux per pole about the end-connections of any phase is 4 = 120 snl ID TttS' Linking with this flux are-- turns, .'. the inductance (interlinkages of flux turns per unit current) per phase is L 4 = j XpX~X 120^p -J- = 30s 2 n 2 The total primary inductance is, then, d , /fo\' , 7.5 s ID 378 ELECTRICAL ENGINEERING Supplying numerical values, s = 4; s' = 2; n = 4; a = 0.95; b = 0.34; d = 0.206; e = 0.244; (-) ' = 1.25; D = 30.46 cm.; p = 6; Z = 11.94 cm.; (&} " = 2.38 \yo/ \{/o/ 10 QQC L = ~io^ [ (0 - 931 + - 843 + L25 + L73)4 + 2 - 38 x 2 J = 0.000249 + 0.0000686 = 0.0003176 henry. Primary leakage reactance is, therefore, z = 27r/Lo = 0.12 ohm Secondary leakage reactance. Self-induction of the secondary is obtained in the same way. Thus the total secondary inductance is Ll a . d where all quantities are known except FIG. 278. In Fig. 278 the leakage path is indicated by the dotted line for the relative position of rotor and stator teeth in which the magnetic conductance is maximum. The reluctance in this case is 0.2625 3.81 + 0.422 = 0.5, and go DESIGN CONSTANTS OF INDUCTION MOTOR 379 Minimum magnetic conductance occurs when the stator and the rotor slots are directly opposite each other. Then 1 p 3 (max.) = 2 ^ + -4- ^ d ~ The value S Q is the magnetic conductance of the small path across the faces of the rotor teeth tips opposite a stator slot. It is, approximately, ' - 5354 - ' 376 ' "' and the average value of the magnetic conductance is /o 2 + 0.376 = -- - - = 1.2 approx. Qo * The rotor self-induction is then determined by supplying numerical values in the equation for LI (see page 370). Thus, 4TX x 11.94 Ll = - - + + " - 4.09 X 10-henry, Reactance = ZirfLi = 0.00154 ohm. Reduced to the primary basis, this gives (9CQ\ 2 -7^-j = 0.0935 ohm. The only quantities remaining to be determined are I m and I h , the magnetizing and core-loss currents. The core loss, or watts lost in hysteresis and eddy currents, is E 2 g, where I h = Eg. This loss is almost entirely in the primary core where the normal frequency of reversal of the flux obtains. The secondary core loss is neglected. 1 As with the alternator, the eddy current loss will be taken as equal to one-half of the hysteresis loss. To find the latter, then, the volume of stator teeth is V t = 72 X 4.23 X 0.45 = 137 cu. in., where 0.45 = area of a tooth. 1 It is neglected only in the case that the pulsation in flux due to the passing of the teeth is negligible, which is, however, not always the case. 380 ELECTRICAL ENGINEERING Weight of teeth = 137 X 0.28 = 38.4 Ib. Hysteresis loss in teeth at 2.8 watts per Ib. at 60 cycles and 64,500 lines per sq. in. is (04 000\ L ffl^jj) 170 watts. Outside diameter of the stator = 16.375 in. Inside diameter of the stator = 14.34 in. Volume of stator core is 14.34 Weight of core = 204 X 0.28 = 57 Ib. Flux density = 516 ' 000 -^ (1.0175 X 4.23) = 60,000 lines per f sq. in. .'. Hysteresis loss in core is 57 X 2.8 X ( 6 ' 1.6 = 140 watts. \64,500y .'. Core loss = 1.5(170 + 140) = 465 watts. 465 Core loss per phase = -y- = 155 watts = E 2 g = EI h , and I h 155 = j=r = 2.44 amp. 244 Per cent. I h = =- = 0.047. Da . That portion of the magnetizing current required by the air gap has been assumed as 20 per cent., or 10.4 amp. That required by the iron part of the flux path must be worked out from the known densities and dimensions of the magnetic circuit. The usual tabulation is as follows: Part Density A.T. per in. Length A.T. Stator teeth Rotor teeth Stator core . . Rotor core . . Total ir< Smax 84,000 72,500 97,000 88,000 60,000 60,000 12 ) 97 7.4( 97 28. Ol 15. 0/ 21 - 5 4.5 4.5 1.15 0.6 4.0 2.0 11 13 18 9 51 . mm max min 3n amp -turns DESIGN CONSTANTS OF INDUCTION MOTOR 381 Since there are eight turns per pole per phase, the magnetizing current required for the iron is calculated from the equation for armature reaction. Arm. reac. = V 2 X 1.5 X amp.-turns per pole and phase = 2.12 X 8 X amp. = 51. .*. Current required for the iron part of the magnetic path = 51 3. 16.96 Total magnetizing current per phase is I m = 10.4 -f 3 = 13.4 per cent. I m = 0.26. Performance curves of the motor may now be worked out from the constants, reduced to percentages, as follows: a 315X52 = 0.0 258 ; = 0.0183; x = 0.0982; Xl = 0.069; / 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 Output (P m -f) FIG. 279. 382 ELECTRICAL ENGINEERING Tabulating as in the problems of the previous chapter, the resulting values are as follows: Slip 0.0 0.01 0.02 0.03 0.05 0.08 0.1 0.3 0.5 1.0 /o 0.257 0.626 1 . 107 1.57 2.38 3.32 3.775 5.34 5.64 5.82 T 0.0 0.498 0.95 1.337 1.925 2.37 2.475 1.675 1.12 0.6 P m -fo 0.0 0.483 0.921 1.288 1.82 2.17 2.217 1.163 0.55 0.0 Pi 0.0462 0.5525 1.022 1.448 2.11 2.675 2.867 2.425 1.953 1.484 E Io 0.257 0.626 1.107 1.57 2.38 3.32 3.775 5.34 5.64 5.82 Eff. 0.0 0.875 0.90 0.889 0.862 0.81 0.772 0.479 0.281 0.0 P.F. 0.18 0.883 0.922 0.921 0.887 0.805 0.76 0.455 0.346 0.255 App. eff. 0.0 0.773 0.83 0.819 0.765 0.652 0.587 0.228 0.0972 0.0 Performance curves are given in Figs. 279 and 280. 2.6 2.4 2.2 2.0 1.8 1.6 1.4 H 1.2 1.0 0.8 0.6 0.4 0.2 t \ 7 cf, 7 0.2 0.4 0.6 0.8 1.0 Speed and Current -7- 4 FIG. 280. 1.2 1.4 1.6 CHAPTER XLIV ROTARY OR SYNCHRONOUS CONVERTERS It is frequently necessary to obtain direct current when the available supply is alternating. Such, for instance, is the case with many electric railways, arc lighting systems, etc. Changing from direct to alternating current is also, but less frequently, required. The work of conversion from direct cur- rent to alternating current or vice versa, is done to great ad- vantage by means of the rotary converter. Looked at from its commutator end, this machine is a direct-current generator; from the other end, it appears to be a synchronous motor with revolving armature. Both receiving and transmitting electrical energy, it is motor and generator combined into one. If it be driven by mechanical, power, it may give out electrical power, in the form of either direct current or alternating current, or both at the same time. Under its various phases of operation, then, the rotary converter may work as a direct-current generator, direct-current motor, synchronous motor or alternator. In any case its performance becomes that of one of these machines or a combination of them, and may be studied in that light. Voltage and Current Ratios. The fundamental equation of a direct-current generator has been found to be E = -r^g- volts, where /, and t, are respectively, the frequency of alternation of voltage in the conductors, the flux per pole cutting the con- ductors and the total number of turns between brushes. If the armature is tapped at two symmetrical points (for each pair of poles), and the taps are brought out to slip rings, then the voltage across the slip rings, by the fundamental equation for alternating-current generators, is E e ff. = -.^ 8 effective volts. This equation holds, however, only for concentrated windings. For distributed windings, it is 2 E'ff. ~ ~ > 383 384 ELECTRICAL ENGINEERING and, substituting for 4.44 its derivation, \/2ir ) the maximum ,- 2 V^TT/V^ 4f4>t voltage is E m = v 2 X - X 1r)8 = -^ volts, which is the same as for direct current. In calculations involving the voltage it is convenient to deal with the voltage to neutral, that is, one-half the direct-current or alternating-current single-phase voltage, and the phase volt age with polyphase connection. Thus, direct-current voltage to E 2ft neutral is E n = -~ = -TT^. Single-phase effective voltage to neutral is E n E V2 2\/2 In a three-phase machine maximum voltage between rings, = \/3 E n = \/3 X Y^- = \/3En. In an TV-phase machine, E m = 2E n sin ^ = E sin - is maximum voltage between rings. Effective voltage is E e /f, = \/2E n siu j=. sin Current relations are determined by assuming 100 per cent, efficiency, or input = output. The direct-current output is 2E n I, where I = direct-current line current. The input is N^I p) for N phases, where I p = phase current. Equating output to input, and solving for I p) 2NEJ P sin -^ 2JU -*-%,-; - 7f 2EJ V%I sin -. N sin ~ Effective line current is the vector sum of two adjacent effective phase currents. This may be proven to be 2V2J ROTARY OR SYNCHRONOUS CONVERTERS 385 Thus, for three-phase, effective phase current is and effective line current is 7 Problem 110. Assuming the direct-current voltage and current to be E and I respectively, deduce effective phase voltages and currents for one-, two-, three-, four-, and six-phase alternating input. For single-phase, the values have already been indicated. They are They cannot be obtained from the general formulae directly, but are easily found from the power equation. The general formulae apply to polyphase machines. Voltage Control. Under the ideal conditions assumed, there is no voltage drop in the armature. Practically, however, there is a drop proportional to the load. With a constant impressed alternating-current voltage on either the rotary terminals or the supply generator, the drop in the armature, or in the armature and line, as the case may be, causes a variation in the terminal direct-current pressure. It is frequently desirable to compensate for this voltage drop in line and armature, in order to maintain constant direct-current voltage across the brushes of the rotary converter. One method of doing this is by proper " compounding of the series field." In general, the procedure is to " under-excite the shunt field," so that at no load the machine, acting as a synchronous motor, will draw lagging current of about one-third normal value. At %-load the current will be FIG 2gl in time-phase with the voltage. At full-load the current, due to the series field m.m.f., will lead somewhat, with the consequent increase in terminal voltage. Heating of the Armature. In Fig. 281 is shown one phase of a rotary converter, whose center line makes an angle (0 a) with the field axis. Let any coil, c, be displaced an angle, a, from the center line of this phase, or 0, from the field axis. 25 386 ELECTRICAL ENGINEERING To Find the Heating in this Coil. Let the current and voltage be in time-phase with each other. Then maximum current will flow in the phase, and therefore in the coil, when 6 a = 0. Thus the current in the coil is not maximum when the coil is on the axis, but when the center of the phase is on the axis. There- fore the alternating current in the coil may be written i = I m cos (6 a). If / is the direct current in the line, ~ is the direct current in 48 the coil. Therefore, the resultant current in the coil is / / lr = l ~~ 2 = COS ' ~~ a ' ~~ 2* But it is known that, in general, 7 1m = N sin ~ Therefore, The heating of the coil is proportional to the square of i r ', the average heating to the average square of the resultant current taken over a half -cycle from = ^to = ^- Thus, average heating of coil = Comparing this heating to that of a direct-current generator with the same current output gives heating of rotary coil _ 1 fi A cos (0 a) x 2 heating of d.-c. gen. coil ~ TT ' ' - l\ d0. Problem 111. Prove that this ratio is: 8 16 cos a 7T + TT N z sin 2 jj NTT sin -^ ROTARY OR SYNCHRONOUS CONVERTERS 387 Proof. /- - _ /i- /4co3(0-_g) _ v J_,\ *- Jl cos 2 (0 - a) - 8AT cos (0 - a) sin ~ + N* sin 2 ^\ ]de 7T - T 2 (l6cos 2 (0-a)-8Arsm^cos (0 - _ 2 Integrating the terms separately, /*- /- / 2 / 2 (1) I 16 cos 2 (6 a)d6 = 16 I (cos 2 6 cos 2 a + 2 cos ^ sin cos a sin a ^~ \ J~\ + sin 2 a sin 2 0) d0 f /sin 26 0\ /- cos 20\ = 16 cos 2 a ( ^ -- h 2 ) + 2 cos a sln a ( - 4 - ) -sin20 = 8r COS 2 a + + Sir sin 2 a =8rr(sin 2 a + cos 2 a) =8*-. (2) j 2 - 8^ sin -^ cos (0 - )d0 '""I = 8JV sin -^ cos a sin + sin a ( cos 0) 2 r sin - cos a. XI r -,* N 2 sin 2 ~ de = \N* sin 2 T? ^ ^ L J ~2 (3) | N 2 sin 2 ^ d0 = | N* sin 2 -^ 9 I* , = TrAT 2 sin 2 ^- J 2 8 16 cos a /. Ratio = - + 1 - - - N 2 sin 2 ^ NTT sin -ft Prove that this ratio is maximum for that coil of the phase for which a = ?r, that is, for the largest possible value of a, and that the ratio is minimum for a = 0. 388 ELECTRICAL ENGINEERING Problem 112. Find the ratio between maximum and minimum heating on three-phase, four-phase, and six-phase rotary converters. So^ion.-Three-phase max. = . O + 1 - ~ = 1.21. 8 16 cos Three-phase mm. = 6QO + 1 - O = 0.226. Three-phase ratio, = = 5.35. 8 16 cos 45 Four-phase max. = 16 sin2 45 o + 1 - 4?r sin 450 = 0.728. Four-phase min. = Jfl g .* 2 450 + 1 - 5^5^ = 0.2. max. 0.728 Four-phase ratio, r = -- = 3.64. 8 16 cos 30 Six-phase max. = 36 gin2 30 <> + 1 - 67rsin30 o = 0.426. Six-phase min. = o + 1 - O = 0.199. ., max. 0.426 Six-phase ratio, = - = 2.14. Problem 113. Find the ratio of average heating around the whole pe- riphery to the heating of a direct-current armature. To determine the ratio of average heating around the periphery the ex- pression must be integrated for the ratio of heating of any coil of a phase for all possible values of a, and this must be divided by the angular width of the phase. Thus, ratio of average heating is N (> a " + 'I i _ 8 16 cos \ x I / - \da r \fr**% ****%) */-- 80 _ 16 sin a "i a _ _ 0.63" N 2 sin 2 - ^ 2 sin 2 - Problem 114. Find the ratio of average heating to direct-current heating for three-phase, four-phase, and six-phase input. Three-phase: + 1 - " - 0.55. Four-phase: - - 0-63 = 0.37. Six -P hase: 36W - ' 63 - - 259 - In all of these problems the input is taken at unity power factor. ROTARY OR SYNCHRONOUS CONVERTERS 389 When there is also a wattless component of current in the winding the heating is due to this component which is not com- pensated for, as well as to the power component. Although the components are at right angles (Fig. 282), the heating due to them is added directly, since it is proportional to their squares, and the sum of the squares gives the total current in the winding. Another way to determine the heating of a rotary converter when the current has a wattless component is to resolve the current into its components and add their instantaneous values. Let the power factor of the alternating current be cos $. Then, if I w is the maximum value of the wattless component of the current in the winding, and I e is the maximum value of the power component, Iw tan = Thus, I w is known, or, I w = I e tan $ = 21 T T* Nsm N . tan <. is positive for leading current. I w is maximum for 6 a 90- -| .'. i w = I w sin (6 a) for a coil displaced a. Thus, the resultant current in a coil, m, is 4 cos (6 a) 4 sin (0 - a) tan < H N sin -JT-- 2v 4 /cos (0 . 7T a) cos + sin (0, a) sin #~7\~ 4 cos < (0 --) i , N sin -JTJ. cos < Problem 116. From this equation determine the maximum heating of an armature coil compared with that of a direct-current generator of the same output. Solution. rotary current in any coil _ ir d.-c. current ~ I_ 2 Ratio of 390 ELECTRICAL ENGINEERING The ratio of heating 4 cos (0 a) __ 1 , 4 sin (0 q) tan \ 16 sin 2 (0 - q) _ 8 cos (0- q) N 2 sin 2 T? N sin - 1 C 2 /16 cos 2 (0 - q) V-I* AT 2 sin 2 ^ 32 cos (0 - ) sin (0 - ) x 8 sin (0 - a) x tan < tan < \d6. AT 2 sin 2 -^ N sin -^ Integrating and applying limits, _ I r___ 4. _i_ 8 TT tan 2 _ 16 cos q 16 sin q tan ) _ 16 cos a + 16 sin tan But 1 + tan 2 = sec 2 ~^ and tan and cos a cos + sin a sin = cos (a ATO .Tn TIT "" N * sin iricos 2 ^ irN sin ^ cos The maximum heating occurs in that coil f on which cos (q 0) is mini- mum. Problem 116. Compare the average heating around the whole periphery with that of a direct-current generator. Solution. For this, it is necessary simply to integrate the above ratio between the limits of a phase, or between jj and +' ?> and divide by ^ 16 cos (a ) x J \da. NTT sin COS r a= ^ ... Ratio = ^ ( : + i 27T y AT 2 sin 2 -^ cos 2 c/=-^ Substituting the expression cos (a 0) = cos cos a sin ^ sin a, and integrating, ^JT 7VY 8a 16 (cos cos a)i a "lv Ratio = =- h r [ tf 2 sin 2 ^ cos 2 NTT sin ^ cos ~ ROTARY OR SYNCHRONOUS CONVERTERS 391 N SX ~N 16 cos < + lir > (2 sin jj \ sin (0) 2ir N 2 sin 2 ^ 8 COS 2 N NTT sin 8 -^cos n AS AT 2 sin 2 -^ cos 2 JV 2 sin 2 -^. cos 2 irN sin - cos ratio = -- 0.63, N* sin 2 r cos 2 where N = 3, 4, 6, and cos = 0.9 and 0.9. The final substitution of values is here left to the student. A good physical idea of the distribution of current in the windings of a rotary armature may be obtained from diagrams showing the direct-current output as a constant fixed in space with reference to the brushes, while the alternating-current input varies according to the angular position of the phases, by the equation i = I m sin 6. Let 6 vary by steps of 30, and write down on the diagrams the values of both the direct and alternating current, with direc- tion arrows, and also the resultant current in each part of the winding. This may be done both for single-phase and three- phase armatures. If the unidirectional line current is 100 amp., then the maxi- mum alternating current is 200 amp. for single-phase and 133 amp. for three-phase. These diagrams are given in Fig. 283. Voltage Control. The voltage of rotary converters for railway work is controlled by variation of the phase of the current input. In general the machine is over-compounded to neutralize the drop in feeders. Therefore, the alternating-current voltage impressed on the rotary, must vary to make up the necessary over-compounding as well as the drop in the armature. 392 ELECTRICAL ENGINEERING The phase of the current, as in a synchronous motor, depends on the field excitation. A reactive coil is placed in series, as shown in Fig. 284. If the current is lagging there will be a big drop in the line and coil, and the voltage impressed on the rotary will be reduced. Current Relations in Single -Phase Converter 30 60 90 100 100 100 Three - Phase Converter 30 60 90 120 FIG. 283. If the series field m.m.f. is so adjusted as to give leading current at full-load, then at light-load, the current will be lagging. By this means of adjustment, a constant direct-current voltage may be maintained. To get this adjustment, the shunt and ROTARY OR SYNCHRONOUS CONVERTERS 393 series field m.m.f. should give unity power factor at about %-load. Question. In a given case, it may be required to obtain con- stant direct-current voltage from a given constant generator voltage, EQ, of such a value as to require the interposition of transformers. How shall the transformer ratio be determined, so as to give the most efficient operation? Line Reactive Coil FIG. 284. If constant voltage, e, impressed on the rotary is assumed, the fundamental equation is Eo = e+ (i + ji')(r + jx) where r -f jx = impedance of line and coil. But generally, e is not constant, but should vary with the load that is, with i. The equation is E Q = e + ik + (i + ji')(r + jx) = e + ik + ir - i'x + j(ix + i'r) (128) where k is the percentage of over-compounding. Assuming that i' = at, say, %-load, the first step is to find i' at no-load, to ascertain that it is not excessive. Let e + ik = e'. Then (128) becomes Eo = e' +(i+ji')(r+jx) E Q 2 = (e + ik + ir - i'x) 2 + (i'r + ix) 2 = e 2 + I 2 Z* + i 2 k 2 -f- 2ei'(& -f r) 2^'x approx. Let i = i'o at non-inductive load. Then E 2 = e 2 + t 2 (Z 2 + k 2 ) + 2ei (k + r) .'. = 1 + i Q 2 1 -h ? (fc -f r) (129) At no-load i = 0. /. E 2 = e 2 + ^ /2 ^ 2 - 2ei f x = e 2 - 2ei'x (approx.) 2ex = ~2xL\T/ "" j' 394 ELECTRICAL ENGINEERING T? The value of is obtained from Eq. 129. Numerical Application. Let e = 1, r = 0.10, x = 0.30, k = 0.10, i = 1 at full-load, and i Q = 0.75. TjJ Then = 1.12 and i' at no-load = - 0.38. 6 The generator voltage should thus at no-load be 12 per cent, higher than the desired voltage at the rotary at no-load, or, in other words, the voltage at the line coming into the substation should be 12 per cent, higher than is actually wanted at the collector rings of the rotary. The rotary will reduce the no- load voltage by taking a large lagging current (in this case 40 per cent, of full-load current). Thus if, for instance, a single-phase 500-volt rotary is to operate from a 10,000-volt generating station, the transformer 1 000 28 2 ratio would not be = 28.2, but = 25.2. If /i is the secondary current due to a certain phase of the primary, whose induced e.m.f. is 6;, then the power component of /i has been shown to be e^ai. jli is evidently the secondary current due to a primary phase which is 90 in space and time behind the former. The induced e.m.f. of this phase is, of course, je^ and the flux causing the e.m.f. is 90 in time ahead of the e.m.f. Thus the flux which reacts on the energy component of the original second- ary current is proportional to j( jei) = ke*. .'. the torque is kdei Voltage Control by Use of the " Split Pole." The " split pole" converter has found extensive application more especially in the field of lighting service, where the fluctuations of the load are not so persistent as they are in railway work. It is based upon the principle that the voltage ratio is altered by shifting the brushes. Ordinarily, the brushes on a direct-current generator are set at or near the neutral point, and when they are in that position, the generator gives maximum voltage. If shifted to any other point, the voltage is E cos a, where a is the angle between the neutral axis and that of the brushes, and E is the voltage when ROTARY OR SYNCHRONOUS CONVERTERS 395 the brushes are on the neutral axis. Under conditions of variable shift, the voltage ratio of a rotary converter is then, E a . c Ecos a FIG. 285. In order to increase the amount of the variation of this ratio, the poles are slotted, or split, directly over the position chosen for the brushes, as shown in Fig. 285, thus enabling the brushes to take a position, which, otherwise, would be impossible owing to bad commutation. Such an arrangement, while it permits a great shift, does not give variable shift. To obtain variation of the shift, which is indeed necessary in order to have voltage control, the poles may be split into two or more separate sections, each with its own exciting coil. Suppose, for instance, an arrangement such as shown in Fig. 286. Let the poles N and S, alone, be excited. The axis of the main flux is then along the line making an angle a with the hori- zontal. Suppose, now, as the load comes on, poles N f and S f are excited by means of a series winding. The flux axis is then shifted toward the horizontal, decreasing a, and consequently increasing the direct- current voltage. The amount of such variation of voltage is considerable. Tests on a certain three-phase machine showed that with constant impressed alternating-current voltage of 194 the direct-current voltage ranged from 317 to 200, giving at all times sparkless commutation. Poor commutation in rotary converters is due almost entirely to self-induction of the coil short-circuited by the brushes. This effect is of sufficient importance to justify the use of inter-poles. Armature Reaction in Rotary Converter Problem 118. Prove that in an ordinary rotary converter the resultant armature reactive m.m.f. due to the direct current and the power component of the alternating current is zero. Solution. Consider a two-pole machine. Let / = direct current, and m = total turns. Then the ampere-turns due to direct current = m X FIG. 286. 396 ELECTRICAL ENGINEERING As these turns are distributed, the direct-current armature reaction = 2 I = m/ This reaction, considering the brushes to be at shift, is along the axis of the brushes. Armature reaction due to the power component of the alternating current, is also along the brush axis, but opposite to the direct-current reaction. Alternating-current armature reaction due to the wattless component is -* /O N/ />"' \S W f\ V * S\ >> a.c. /N * I where i' a . e . I a .c. sin a, and T = effective turns per phase. With the power component, the expression for armature reaction is exactly the same but the direction of the reaction is along the brushes in- stead of along the field axis as with the wattless component. .'. Alternating-current armature reaction = N V2 i a . c . T where i a . c . is the power component, in the direction of the direct current. But V2/ l>a.c. Substituting this ^ V2l a.c. To = IT, Effective turns per phase, T = kt, where t chord N k = ratio, turns per phase. arc 7 sln F To = t sin -. Also, Nt = m. Substituting these values, alternating-current armature reaction due to i = -5- = direct-cur- rent armature reaction, and the total reaction in the brush axis is 0. Armature reaction is present in split-pole ro- taries to a limited extent. The direct-current reaction is always along the brush axis, while the alternating-current reaction due to the power component of the current is at right angles to the resultant flux. The angle, a, Fig. 286, is the angle of relative brush shift. The direct- current reaction, F, Fig. 287, may then be resolved into components, F\ along the flux axis, and Ft in line with the alternating-current reaction. The alternating-current reaction due to the power component of current is in line with F 2 , and is F n = F cos a. .'. F 2 = F cos a = F n . Pole Axis FIG. 287. ROTARY OR SYNCHRONOUS CONVERTERS 397 Also, Fsin F n cos a sin a = F n tan a. Thus, F z is compensated by F n , while FI has no compensation except in so far as this is accomplished along the lines of the main field excitation. There remains an uncompensated component F 3j along the brush axis, which is F, = IP, sin cos a Example. Let .a = 30. Then sin a = sin 30 = 0.5. sin 2 a = 0.25. cos a = 0.866. 0.25 F n X 0.866 0.29F n . That is, the uncompensated armature reaction amounts to about 30 per cent, of the alternating-current reaction. Transformer Connections for Rotary Converters. Most rotary converters receive energy from three-phase supply. In most cases it is simply a matter of connecting to the transformer terminals as would be done in the case of a synchronous motor. However, it is often economical to operate the rotaries as six- phase machines, receiving the energy, however, as three-phase. In connecting six-phase, the principle is to always have the direction arrows as shown in Fig. 288, for either A or Y-connection. This is illustrated in Fig. 289 by what is called the double A-connection. Each primary supplies power for two secondaries whose terminals are led to the slip rings of 2,5 FIG. 288. i ,1*35?") 12 11 10 9 Sec. FIG. 289. FIG. 290. the rotary converter. There are six slip rings. Each slip ring is connected to two transformers. The terminals are indicated by corresponding numbers. Fig. 290 illustrates the double T-con- nection. The arrows show the only essential precaution that is necessary to take. There are still other ways of connecting six- phase, as the "diametrical," in which the terminals of each 398 ELECTRICAL ENGINEERING transformer are connected to diametrically opposite points on the armature. Synchronous Condensers. When a rotary converter or syn- chronous motor is over-excited so that it takes a leading current, it may be used on a system for the purpose of improving the power factor. Such uses are of frequent occurrence where the load consists principally of induction motors with their large lagging compo- nents of current. Machines used for the purpose of neutralizing this lagging current by drawing on the supply for an equal leading current, are called "synchronous condensers." The commercial problem is, in any case, to determine whether the saving in line and generator loss, improvement in regulation, and, with initial installations, the saving in capital expenditure, justify the additional expenditure required for the installation of synchronous condensers. As a concrete illustration, consider a system with poor voltage regulation. Can the owners afford to buy synchronous con- densers, at say, $10 per kv.a., in order to improve the operation of the system? Let it be assumed that the power factor is ordinarily only 70 per cent., that the cost of energy to the station is Ic. per kw. hr., and that the load factor is 30 per cent. At normal full-load, i = 1 = load component of the current. Then, since cos = 0.7, i f = 1 = wattless component, lagging, and the total current is / = i ji' = 1 jl. To neutralize the lagging component, a leading component, i' c = 1, is required. If this is the full-load current of the condenser, its rating is at once determined and, thereby, its cost. It would probably be better not to try to make the current lead by 90. For instance, let the condenser also do some work, say i' c = 0.1. Then Ic = 0.1 +jl, and I c = lM = 1-005. Thus the condenser current is practically unaffected in amount by the addition of a 10 per cent. load. The total station load is now i + i e = 1.1. Whereas, pre- viously its current was / = 1.4, it is now only 1.1, and yet the useful load is not only the same but is 10 per cent, greater. The voltage of the load is taken as e = 1. ROTARY OR SYNCHRONOUS CONVERTERS 399 The line loss has been reduced in the ratio ( , or 38 per cent. Assuming, previously, a line and generator loss of 20 per cent. which would be reasonable under the conditions, the saving amounts to 0.38 X 0.2 = 0.076 watt. At 30 per cent, load factor, this saving is 0.076 X 0.3 = 0.0228. The gain in power from the condenser is 0.1 X 0.3 = 0.03. At Ic. per kw. hr., and assuming 7200 hr. per year, the value of energy saved is $0.00001 X 7200 X 0.0528 = $0.0038 Cost of condenser at $10 per kv.a. is $0.01 X 1.0 = $0.01 Interest at 6 per cent, on cost of condenser = 0.01 X 0.06 = $0.0006. Problem 119. Apply the above reasoning to a system in which the normal load current is 1000 amp. at 250 volts, stating the conclusion with reference to the advisability of buying synchronous condensers. Considering the line to have resistance only, show how the voltage regulation would be affected by the change. There is still one feature of the use of synchronous condensers that has not been considered. In the above example, the load has been taken as varying to give a load factor of 30 per cent., while the synchronous condenser was assumed to supply at all times a leading component exactly neutralizing the lagging component of the load. This assumption of flexibility of the condenser is hardly justified. At the same time it would be undesirable that the condenser should draw full-load leading current continuously. The field excitation may therefore be obtained by compound winding such that at full-load i' c = 1, while at no-load i' e = 1. The field of the condenser must therefore be made to depend on the entire load to be compensated. By making the full-load and no-load wattless components equal and opposite, the smallest condenser is required. This gives, however, consider- able no-load line loss. The use of synchronous condensers has now been discussed sufficiently to enable the student to investigate any given case and make an engineering report on it. The question of where to install the condensers, whether at the load, or at the power station, is also of interest, and should be discussed by the student, reasons being given why either position is to be preferred. CHAPTER XLV SINGLE-PHASE ALTERNATING-CURRENT MOTORS Under this heading would naturally be comprised single- phase induction motors and the various types of commutator motors. The development of the latter class of machines, to- gether with certain inherent defects in the former, has had the effect of rendering the single-phase induction motor nearly obsolete. When a polyphase induction motor is running, if one of the phase circuits be opened the motor will continue to operate as a single-phase machine. Its permissible output will be greatly reduced and, in general, its characteristics will be changed for the worse. The principal difficulty with the single-phase induction motor is its inability to start. Special means have to be supplied, such as "splitting" the phase, that is, dividing the primary wind- ing into two circuits, one of which is provided with a condenser or resistance to give a time-phase displacement of one current rela- tive to the other. In some such way the motor is temporarily converted into a poor polyphase motor with just sufficient torque to enable it to start without load. After coming up to speed, a switch is operated which causes the motor to run thereafter through power supplied to one phase. The Series Motor. It has already been pointed out that a direct-current series motor possesses, fundamentally, the quali- fications for operation on alternating current. Practically, in direct-current motors, the field is made strong and the armature weak. In alternating- current series motors the reverse is the case; the armature m.m.f. is three or FIG. 291. four times as strong as that of the field circuit. To determine the values of flux, current, torque, power, etc., we may proceed as follows: Consider an armature coil displaced 6 from the horizontal (Fig. 291). The flux enclosed by the coil is = m sin co sin 400 ALTERNATING-CURRENT MOTORS 401 since the flux is alternating, the total flux at any instant being ra sin co. Therefore the induced e.m.f. per coil is d / d6\ de = rr = 3?m ( w cos w sin 6 + sm w cos -77 ) at \ ail Let, now, B = co^, where o>i = Zirfi and/i = frequency of rota- tion, i.e., due to rotation of the armature; the coil has moved through an angle B in the time t, wi being the angular velocity of the rotation. Then dB _ dt "" Wl ' and de 3> m (a> cos ut sin B + wi sin co cos B). T If there are T turns between brushes, there are dB turns in the little element dB. Therefore the total induced e.m.f. is T C Q= 2^ e SB I ,n(a> cos co sin B + coi sin co cos 0) *Jt sn co^ volts - Thus, the induced e.m.f. is seen to be of fundamental frequency I, and in time-phase with the flux. Let the components of the flux be a , the armature flux along the brush axis. These component fluxes are then in space quadrature. Let, also, N/ be the number of field turns per pole, and N a the equivalent 2 number of armature turns (N a = -T for distributed winding). If the magnetic reluctance were uniform about the armature periphery, -f would be equal to -r/- 1. 26 402 ELECTRICAL ENGINEERING Let Nf = n iVa where n < 1. Then $/ = mn. Since, now, the conductors are rotating in a magnetic field, there is induced in them an e.m.f., E r , of rotation, whose effective value is N a , here, is used for the effective turns of an equivalent single- phase alternator. The induced e.m.f. in the armature due to the alternation of the armature flux, < a , is VOltS, and, similarly, the induced e.m.f. in the field due to the alterna- tion of the field flux, a and $/. .'. the total induced voltage is E t = v E r z -\- (E a -\- Ef) 2 . But, E r flN a S E f fNf n where s = j, and, as before, n = j/* Also, E a Ef~ N a $ a 1 Nf$f ~ mn 2 ' .*. Substituting n E r = -E f and E a = &y n ' mn* E f I (ran 2 + I) 2 E f - ALTERNATING-CURRENT MOTORS 403 If resistance drops in the field and armature coils are neglected, Ei = E, the impressed e.m.f. The reactance may be determined as in the case of the induction motor. Let x/ = reactance of pi field winding. Then xj -/ Substituting in the equation for induced e.m.f., En I = - 7=- = - --= approx. The electrical power input is Pi = El cos a. Mechanical power output is P- FT - En r EJ T L -p Torque = T = j synchronous watts. S torque at synchronous speed torque at standstill T. Ao (mn 2 + I) 2 = T Q ~~ A 8 ~ mV + (mn 2 + I) 2 ' As an example, assume a motor of uniform reluctance (m 1), and of the same number of turns in the armature and field coils (n = 1). Then T 2 2 4 TO = r+~2" 2 = 5 = - 8 - In such a machine the starting torque is not much greater than that of full-load, which is not very satisfactory for the class of work series motors are usually called on to perform. An approximate ratio, -f Using this expression, if n = 0.5, The starting torque is five times as strong as that at synchron- ous speed. 404 ELECTRICAL ENGINEERING Compensated Series Motor. The practical operation of the series motor is attended with difficulty owing to the tendency to excessive sparking. It also has poor power factor. In the short-circuited coils heavy electromotive forces are produced by the alternating flux. To overcome these electromotive forces compensating windings are supplied which under speed conditions act like inter-poles to neutralize the armature reaction and self-induction of the short- circuited coils. Fig. 292 gives the wiring diagram of the com- pensated series motor. Problem 120. Derive the complex equations for the series alternating- current motor with compensating coil. Solution. The impressed e.m.f. may be written E = IR +E r +_j(Ix + E f \ where R = total resistance of armature, field and compensating coil, x self -inductive reactance of armature and coil, E r = e.m.f. due to rotation, Ef = induced e.m.f. in the field. It has been shown that also Substituting, 1JT Er E \/2 0/, and be resolved into two components, fa = cos in the direction of the brush axis and t would cease to exist. In that case there would be no xl drop across the " transformer " poles, and if the rl drop were neglected, the whole impressed voltage would be across the "field" poles. Denoting by E the impressed e.m.f., by E t the drop across the transformer poles, and by E f the drop across the field poles, then, assuming perfect mutual induction at standstill, E t = 0, and E f = E. FIG. 295. 408 ELECTRICAL ENGINEERING As the armature starts to revolve due to the torque between the field flux and current, an e.m.f. is generated in it by rotation. This e.m.f., E r , is in time-phase with the flux $/, and generates current, / r , in the short-circuited armature. / r must attain a value such that L-r E r . L-r may be called the e.m.f. of alternation of the current. It has been assumed that at standstill t to equal f (inst.) = $/ cos 6, and of fa, is t (inst.) =!>/ sin 6. turns on transformer poles Nt Let n turns on field poles N/ It must be remembered that this machine is hypothetical; there are not actually two sets of poles. The relative values of N t and N/ depend entirely on the brush shift. At 45 shift they are equal, and n = 1. At 75, or a = 15, n > 1. The e.m.f. of rotation, then, evidently depends on the speed, the brush shift, and E f) the voltage across the field poles. It is E r = snE f . These two e.m.fs., that of the field and that due to rotation of the armature, make up the total e.m.f., E. Since they are in quadrature, E = \/E r 2 + E f 2 = E f \/n 2 s 2 + 1. ALTERNATING-CURRENT MOTORS 409 The current, since rl drop is neglected, is I = &. X f On the basis of perfect mutual induction, also, the transformer induced armature ampere-turns = ampere-turns on the trans- former poles, or I t N a = IN t . where I t is the current in the armature due to the inductive action of the transformer poles. Also, the current in the armature due to E r gives IrNa = J INf, that is, the armature ampere-turns due to rotation are propor- tional to the ampere-turns on the field poles in the ratio -j* These two currents, I t and / r , are evidently in time-quadrature although in space-phase. The total armature current is, therefore, /. = /. ITU Power factor of the motor is determined from the components of the impressed e.m.f., E. Thus, r E ' power factor = cos a = -~- Power developed is P = E r l. p Torque = r- j- synchronous speed Current input, "fjl T7f f[j f f*j x f x f \/n 2 s 2 -f 1 Where imperfect mutual induction exists it is necessary to introduce the term x t to account for the self-induction of the transformer poles, INDEX Abampere, 1, 21 Abohxn, 1 Abvolt, 1 Admittance, 110 Air gap of alternators, 293 of induction motors, 365 Alternating current, 38 Alternator, 246 definite pole, 258 design, 289 field winding, 298 heating, 299 rating, 250 reactance, 270 regulation, 300 round rotor, 259 short-circuit, 301 three-phase, 246 two-phase, 246 Ampere, 1, 21 Apparent efficiency of transmission, 117 Armature, drum, 48 laminations, 296 length, 58, 294 reaction, 47, 60, 252, 278, 322, 395 resistance, 64, 295 ring, 48 winding, 46 Auto-transformer, 240 for two-p has e three-p h a s e transformation, 241 with induction motor, 364 Average value of sine wave, 41 Balancer, 90 Ballistic galvanometer, 102 Bar windings, inductance of, 275 Battery for three-wire system, 90 Belt leakage flux, 374 Booster, 94 British thermal unit, 7 BROOKS and TURNER, 36 Brush design, 66 resistance, 65 Calorie, 7 Capacity, 123 distributed, 162 of a concentric cylinder, 155 of a sphere, 124, 153 of a spherical concentric con- denser, 154 of a three-phase cable, 158 of a transmission line, 156 of two parallel plates, 156 reactance, 126 Characteristic, motor speed, 99 torque, 100 Charge of a condenser, 123 Charging current of a transmission line, 158 Circuit, magnetic, 57 Coercive force, 19 Commutating pole, 74, 82 winding, 80 Commutation, 82 Commutator, 66 Compensated A.-C. series motor, 404 Compensating winding 74, 80, 82. Compensators; see Auto- trans- formers. Complex quantities, addition of, 169 differentiation of, 172 division of, 170 exponential representation of, 172 involution and evolution of, 170 logarithm of, 172 multiplication of, 170 representation of, 169 roots of, 171 Compounding curves of alternators, 266 Condenser, 123 synchronous, 396 411 412 INDEX Conductance, 2, 110 Conductivity, 2 Constant-current transformer, 121 potential constant-current transformation, 120, 140 Core-loss current of induction motor, 379 current of transformer, 176 dependence on e.m.f. wave shape, 194 relation to form factor, 195 Cosine series, 171 Cost of transformers, 223 Coulomb, 1, 15 Cross-magnetization, 49 Current, 1 density, 56 / in alternator armature, 291 distribution in rotary con- verters, 391 ratios in rotary converters, 383 Cylindrical poles, 15 "Dead points" in induction motor, 369 Demagnetization, 49 Density, current, 56 energy, 17 of magnetic field, 14 Design of alternators, 289 of D.-C. generators, 55, 74 of induction motors, 365 lifting magnets, 26 Dielectric strength, 125 Distorted waves, 133, 196, 232 Distributed capacity, 162 inductance, 162 winding, 50 three-phase, 281 Distribution factor of three-phase winding, 292 DOBROWOLSKY, 91 Drum winding, 48 Ducts, ventilating, 58, 294 Eddy current loss, 187 of D.-C. generator, 70 in transformers, 211 Edison three-wire system, 89 Effectiveness of coil, 42 Effective value, of distorted wave, 134 of sine wave, 40 Efficiency of D.-C. generator, 68 of transformer, 210, 220 of transmission, 117 Electro-dynamometer, 146 E.m.f. waves, generation of, 30 End connections, inductance of, 271, 377 stresses on, 319 Energy density, 17 of short-circuit, 314 Equalizer, 89 Exciting current of transformer, 174, 191 Exponential series, 172 Farad, 157 FARADAY, 14, 29, 44, 124 Field current at short-circuit, 323 voltage at short-circuit, 323 winding of alternators, 298 Flat poles, 15 Flux calculation, 56 for alternators, 292 density in teeth, 294, 367 in transformers, 212 distribution around armature, 67 leakage, 56 Fly-wheel effect on hunting, 287 Force, lines of, 14, 124 on wire in a field, 21 tubes of, 124 Form factor, 42 relation of core loss to, 195 FOURIER'S series, 190, 196 FRANKLIN, 124 Frequency, 38 Friction loss in D.-C. generators, 70 Fringing factor, 58 FROELICH'S equation, 51 Galvanometer, ballistic, 102 as ammeter, 103 GAUSS, 15 GAUSS' theorem, 15 INDEX 413 Generator, alternating-current, 246 design of a D.-C., 55 direct-current, 45 efficiency of a D.-C., 68 homopolar, 44 induction, 258 losses in D.-C., 68 three-wire, 90 turbo-, 258 Generators in parallel, 88 in series, 89 Gradient, potential, 153 Heating of alternators, 299 of D.-C. generators, 72 of rotary converters, 385 of transformers, 222 Henry, the, 33 Horoopolar generator, 44 Horsepower, 100 of induction motor, 348 Hunting, 283 Hysteresis, 18 constants, 187 loop, 186, 190 loss, 186 in D.-C. generator, 70 in transformers, 211 Impedance, 105, 108 condensive, 126 triangle, 106 Inductance, 33 distributed, 162 of bar windings, 275 of concentric cable, 159 of end connections, 271, 377 of transformers, 202 of transmission line, 119, 160 maximum, of coil, 36 Induction motor, 342 abnormal operation of, 361 air gap, 365 at end of transmission line, 364 core loss, 380 current, 379 "dead points," 369 design, 365 equivalent circuit, 346 Induction motor, horsepower of, 348 magnetizing current, 379 maximum output, 349 performance curves, 349, 381 power factor of, 348 reactance, 373 secondary resistance, 352 rotary field, 343 single-phase, 400 slip of, 345 slot and tooth dimensions, 366 squirrel-cage winding, 362 theory of operation, 344 torque of, 347 types of, 362 with auto-transformer, 364 mutual, 117 Inductive circuit, characteristic fea- tures of, 39 current in, 35 fundamental equation of, 38 Instruments, formulae of magnetic, 24 Insulation in slots, 291 thickness in transformers, 215 Intensity of electric field, 152 of magnetic field, 14 Intel-linkage factor, 159 Joule, the, 6 KENNELLY, 113 KIRCHOPF'S laws, 4 Laminations, 296 Leakage factor, 77 flux, 56 Lifting magnet, 26 Lines of force, 14, 124 Losses in D.-C. generator, 68 Magnet, lifting, 26 for metors, 16 pull of, 16 Magnetic circuit, 57 dimensions, 296 cycle, 17 density, 14 field intensity, 14 414 INDEX Magnetic intensity determinations of, 21 , Magnetism, 14 molecular theory of, 17 residual, 17 Magnetization, 18 curves, 26 Magnetizing current of induction motor, 379 of transformers, 176, 213, 220 Magnetomotive force, 18 Maximum output of induction motor, 349 short-circuit current, 305 Molecular theory of magnetism, 17 Motor, A.-C. single-phase, 400 D.-C., 98 induction, 342 principle of, 20 repulsion, 407 speed characteristics, 99 synchronous, 324 WlNTER-ElCHBERG, 407 Multiphase short-circuits, 320 Mutual induction, coefficient of, 177 Natural period of a machine, 284 Network, solution of, 4 OERSTED, 20 Ohm, 1 OHM'S law, 3 ONNES, 301 Open delta transformer connection, 236 Parallel circuit calculations. 129 Permeability, 14, 19 Phase characteristics of synchron- ous motors, 335 difference, 106 Pole intensity, 17 Potential, 162 difference, 1 gradient, 153 Power, 6, 42 average value of, 114 by symbolic method, 113 equation, 35 Power factor, 116 of A.-C. generator, 117 of D.-C. motors, 99 of short-circuit, 305, 318, 321 of three-phase alternator, 249 Pull of magnet, 16 Rating of alternators, 250 of auto-transformers, 240 of T-connected transformers, 238 Reactance, 39 of alternators, 270 of armature coils, 258 of induction motors, 373 of synchronous motor, 324 of transformers, 221 Regulation of alternators, 300 of transformers, 221 - of transmission line, 117 Reluctance, 68 of three-phase transformers, 234 Repulsion motor, 407 Residual magnetism, 17 Resistance, 1 armature, 64 brush, 65 in induction motor secondary, 352 of rotor, 371 of synchronous motor, 324 series field, 65 shunt field, 62 Resistances in parallel, 3 in series, 3 Resistivity, 2 of conductors, 3 Resonance, 128 effects, 137 Resultant field of three-phase sys- tem, 254 Ring winding, 48 Rotary converter, 383 armature reaction, 395 current distribution in, 391 heating of, 385 six-phase, 385, 397 transformer connections for, 397 INDEX 415 Rotary converter, voltage control, 385, 391 voltage and current ratios, 383 with split poles, 394 field, 343 Rotor resistance calculation, 371 slots, number of, 369 Saturation curves, 26 calculation of, 59, 78 SCOTT connection, 237 Series circuit calculation, 126 field winding, 63 lighting circuits, 120 motor, A.-C., 400 starting torque of, 403 compensated A.-C., 404 Short-circuit current, maximum, 305 energy of, 314 field current of, 323 field e.m.f. of, 323 multiphase, 320 of Uternators, 301 power of, 305, 318, 321 stress on shaft due to, 313 Shunt field winding, 62 SIEMENS'S e 1 e c t r o-dynamometer, 146 Sine series, 171 wave, average value of, 39 effective value of, 41 Single-phase A.-C. motors, 400 Six-phase rotary converter, 385, 397 Slip of induction motors, 345 Slot design, 276 dimensions of alternators, 290 of induction motors, 366, 369 insulation, 291 pitch, 290 Slots, number of rotor, 369 Speed characteristics of motors, 99 Split pole rotary converters, 394 Squirrel-cage winding, 362 Starting torque of series A.-C. motor, 403 Stator teeth flux density, 367 STEINMETZ, 113, 162, 178, 187, 198 Stresses in transformers, 202 calculation of, 206 Stresses on armature end connec- tions, 319 Susceptance, 110 Symbolic method, 113 Synchronous condensers, 398 converters; see Rotary con- verters. impedance, 302 motor, 324 diagram, 325 equations, 327 phase characteristics, 335 reactance, 324 resistance, 324 V-curves, 326, 341 watts, 348 Teeth flux density, 294, 367 Temperature coefficient, 2 of conductors, 3 THOMSON, ELIHU, 407 Three-phase system, 227 resultant field of, 254 wave distortion in, 230 with neutral, 228 transformers, 229, 233 reluctance of, 234 Three-wire generator, 90 system, 89 Torque characteristics, 100 of D.-C. motors, 99 of induction motors, 347 "Transformer action" in single- phase motors, 402 Transformer, auto-, 240 calculation, example of, 179 approximate method of, 181 coil connections, 216 connections for rotary con- verters, 397 core area, 214 length, 214 loss current, 176 diagram, 175 design, 209 efficiency, 210, 220 equivalent circuit, 178 exciting current, 174 flux calculation, 214 416 INDEX Transformer, flux density, 212 heating, 222 inductance, 202 insulation thickness, 215 losses, 211 magnetizing current, 176, 213, 220 number of turns of, 214 rating, 179 reactance, 221 regulation, 221 weight, 223 winding calculations, 215 Transformers connected open delta, 236 three-phase, 244 core type, 209 cost of, 223 cruciform type, 209 for six phases, 245 in parallel, 243 in series, 242 lighting, 211 methods of cooling, 210 power, 211 rating of T-connected, 238 shell type, 209 stresses in, 202 calculation of, 206 T-connected, 237 Transmission line calculation, 113, 165 approximate, 132 Transmission line capacity, 156 Tubes of force, 124 Two-phase three-phase transforma- tion, 238 Unit pole, 14 Units, 1 V-curves of synchronous motor, 326, 341 Vector, see Complex quantities. addition, 113 multiplication, 113 Ventilating ducts, 2.94 Volt, 1 Voltage control of rotary converters, 385, 391 ratios of rotary converters, 383 Wattless component, 108, 110, 145 Wattmeter, 146 compensated, 148 connections, 147 correction factor, 150 errors, 150 Wave analysis, 196 distortion in transformers, 189 Waves, generation of e.m.f., 30 Weight of transformers, 223 WiNTER-EiCHBERG motor, 407 Zero vector, 108 "Zig-zag" flux, 374 A. TT TNI**-** i i IRE TO nc.'*""- UNIVERSITY OF CALIFORNIA LIBRARY *&&(